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forward and reverse reactions continue at identical rates, so there is no overall change in the amounts of reactants and products."},{"text":"exceeds","correct":false,"explanation":"A faster forward rate would cause a net conversion of reactants into products. At dynamic equilibrium, the forward reaction rate equals the reverse reaction rate."},{"text":"is less than","correct":false,"explanation":"A slower forward rate would produce a net conversion of products into reactants. At dynamic equilibrium, the forward and reverse reaction rates are equal."},{"text":"alternates with","correct":false,"explanation":"The reactions occur simultaneously rather than taking turns. Dynamic equilibrium requires the forward reaction rate to equal the reverse reaction rate."}]},{"id":"e1c7ff24-fdf5-43ed-8bac-dbdbb060a82b","cardType":"basic","points":2,"front":"Why must an equilibrium reaction be in a closed system?","back":"So no substances enter or leave, allowing equilibrium to be maintained.","mcqOptions":[{"text":"So no substances enter or leave, allowing equilibrium to be maintained.","correct":true,"explanation":"A closed system prevents matter from being added or lost. This allows the forward and reverse reaction rates to become equal and the equilibrium composition to remain constant."},{"text":"So no energy can enter or leave, allowing equilibrium to be maintained.","correct":false,"explanation":"A closed system can exchange energy with its surroundings, so constant energy is not its defining requirement. No substances may enter or leave if equilibrium is to be maintained."},{"text":"So no catalysts enter or leave, allowing equilibrium to be maintained.","correct":false,"explanation":"Catalysts may be present in an equilibrium system and do not determine whether it is closed. A closed system prevents substances from entering or leaving."},{"text":"So all reactions completely stop, allowing equilibrium to be maintained.","correct":false,"explanation":"Reaction rates remain non-zero at dynamic equilibrium, so reactions have not stopped. The system must be closed so no substances enter or leave and disturb equilibrium."}]},{"id":"42598673-e8fd-4f06-a967-ab2f481e32a8","cardType":"basic","points":1,"front":"At dynamic equilibrium, what happens to reactant and product concentrations?","back":"They remain constant, but not necessarily equal.","mcqOptions":[{"text":"They remain constant, but not necessarily equal.","correct":true,"explanation":"Equal forward and reverse reaction rates cause reactants and products to be consumed and formed equally quickly. Their concentrations therefore remain constant, although their values need not match."},{"text":"They become equal, and necessarily constant.","correct":false,"explanation":"Equilibrium does not require equal concentrations; it requires equal forward and reverse reaction rates. Consequently, both concentrations remain constant but may have different values."},{"text":"They both decrease, but not necessarily equally.","correct":false,"explanation":"Concentrations do not continually decrease because each substance is replenished by the opposing reaction. At equilibrium, reactant and product concentrations remain constant but need not be equal."},{"text":"They both increase, but not necessarily equally.","correct":false,"explanation":"Concentrations do not continually increase in a closed equilibrium system. Equal forward and reverse rates keep them constant, although reactant and product concentrations may differ."}]},{"id":"af0c0200-e754-4b55-a80e-838cf2166db9","cardType":"basic","points":2,"front":"If more reactant is added to an equilibrium mixture, which way does equilibrium shift?","back":"Towards products, using up some added reactant.","mcqOptions":[{"text":"Towards products, using up some added reactant.","correct":true,"explanation":"Adding reactant disturbs equilibrium by increasing its concentration. The system responds by favouring the forward reaction, forming more products and using up some added reactant."},{"text":"Towards reactants, producing more added reactant.","correct":false,"explanation":"Shifting towards reactants would produce still more reactant rather than oppose the concentration increase. Equilibrium shifts towards products to consume some of the added reactant."},{"text":"Towards neither side, leaving all added reactant.","correct":false,"explanation":"The equilibrium position changes when a reacting substance is added. It shifts towards products, consuming some added reactant until a new equilibrium composition is established."},{"text":"Towards both sides, using up all added reactant.","correct":false,"explanation":"Equilibrium reactions always continue in both directions, but a disturbance creates a net shift in one direction. Adding reactant causes a net shift towards products."}]},{"id":"d15886f7-8825-4afc-8cd3-e19b28959a15","cardType":"basic","points":2,"front":"If a product is removed from an equilibrium mixture, which way does equilibrium shift?","back":"Towards products, replacing some removed product.","mcqOptions":[{"text":"Towards products, replacing some removed product.","correct":true,"explanation":"Removing product lowers its concentration, so the reverse rate initially decreases. The forward reaction is favoured overall, producing more product until equilibrium is re-established."},{"text":"Towards reactants, replacing some removed product.","correct":false,"explanation":"A shift towards reactants would consume more product and reinforce its removal. The equilibrium instead shifts towards products, replacing some of the product that was removed."},{"text":"Towards neither side, leaving the product depleted.","correct":false,"explanation":"Removing a reacting substance disturbs the equilibrium composition, so the position does change. It shifts towards products to replace some of the material removed."},{"text":"Towards both sides, replacing all removed product.","correct":false,"explanation":"The equilibrium shift replaces only some removed product because the equilibrium constant remains unchanged at constant temperature. The net shift is towards products, not equally in both directions."}]},{"id":"2416ef85-cb48-4da4-9f92-38d300817a34","cardType":"cloze","points":null,"front":"Pressure significantly changes equilibrium position only when [ ] are involved.","back":"Pressure significantly changes equilibrium position only when gases are involved.","mcqOptions":[{"text":"gases","correct":true,"explanation":"Pressure changes affect equilibria involving gases because gas concentrations and partial pressures change when volume changes. The favoured side depends on the numbers of gaseous moles."},{"text":"solids","correct":false,"explanation":"Pure solids are essentially incompressible, so changing pressure does not significantly alter their concentrations. Pressure significantly affects equilibrium position when gases are involved."},{"text":"liquids","correct":false,"explanation":"Liquids are nearly incompressible, so ordinary pressure changes have little effect on their concentrations. Significant pressure effects on equilibrium require gaseous substances."},{"text":"catalysts","correct":false,"explanation":"A catalyst changes both forward and reverse rates equally and does not alter equilibrium position. Pressure has a significant effect only for equilibria involving gases."}]},{"id":"f52597f6-b8d9-4543-9644-3a97868cfa35","cardType":"basic","points":1,"front":"Increasing pressure shifts a gaseous equilibrium towards which side?","back":"The side with fewer moles of gas.","mcqOptions":[{"text":"The side with fewer moles of gas.","correct":true,"explanation":"Increasing pressure favours the equilibrium side occupying less gaseous volume. At the same conditions, this is the side containing fewer total moles of gas."},{"text":"The side with more moles of gas.","correct":false,"explanation":"The side with more gaseous moles would occupy a greater volume and is favoured by lower pressure. Increasing pressure shifts equilibrium towards fewer moles of gas."},{"text":"The side with reactant gas molecules.","correct":false,"explanation":"Pressure does not automatically favour reactants; the direction depends on gaseous stoichiometric amounts. Increasing pressure favours whichever side has fewer moles of gas."},{"text":"The side with product gas molecules.","correct":false,"explanation":"Pressure does not automatically favour products; products are favoured only if they contain fewer gaseous moles. The correct criterion is the total gas moles on each side."}]},{"id":"ccde432b-d439-4bc5-81ce-aab291d6f8af","cardType":"basic","points":1,"front":"If both sides have the same number of gas moles, how does pressure affect equilibrium position?","back":"No effect.","mcqOptions":[{"text":"No effect.","correct":true,"explanation":"When both sides contain the same total number of gaseous moles, compression changes their pressures proportionally. Neither direction reduces gas moles, so the equilibrium position is unaffected."},{"text":"Shifts towards products.","correct":false,"explanation":"Increasing pressure favours products only when the product side has fewer gaseous moles. Equal gas moles on both sides mean pressure has no effect on equilibrium position."},{"text":"Shifts towards reactants.","correct":false,"explanation":"Increasing pressure favours reactants only when the reactant side has fewer gaseous moles. If both sides have equal gas moles, there is no equilibrium shift."},{"text":"Stops both reactions.","correct":false,"explanation":"Pressure does not stop the forward and reverse reactions, which continue dynamically. With equal gaseous moles on both sides, pressure has no effect on equilibrium position."}]},{"id":"f5933bcb-02b8-4786-a1a9-5ac92dd69b07","cardType":"basic","points":1,"front":"For an exothermic forward reaction, what happens to $K_c$ when temperature increases?","back":"$K_c$ decreases.","mcqOptions":[{"text":"$K_c$ decreases.","correct":true,"explanation":"Increasing temperature favours the endothermic reverse reaction. This reduces the equilibrium product-to-reactant concentration ratio, so the equilibrium constant for the exothermic forward reaction decreases."},{"text":"$K_c$ increases.","correct":false,"explanation":"An increase would occur for an endothermic forward reaction. Here, higher temperature favours the endothermic reverse reaction, so the equilibrium product-to-reactant ratio falls and $K_c$ decreases."},{"text":"$K_c$ stays constant.","correct":false,"explanation":"Although concentration and pressure changes leave the equilibrium constant unchanged, temperature does not. Heating favours the endothermic reverse reaction, so $K_c$ decreases."},{"text":"$K_c$ becomes zero.","correct":false,"explanation":"Heating shifts the equilibrium towards reactants but does not eliminate all products, so the equilibrium constant does not become zero. For an exothermic forward reaction, $K_c$ decreases."}]},{"id":"fee61014-188a-41a9-b383-817788811cd6","cardType":"basic","points":1,"front":"For an endothermic forward reaction, what happens to $K_c$ when temperature increases?","back":"$K_c$ increases.","mcqOptions":[{"text":"$K_c$ increases.","correct":true,"explanation":"Increasing temperature favours the endothermic forward reaction. This raises the equilibrium product-to-reactant concentration ratio, so the equilibrium constant for the forward reaction increases."},{"text":"$K_c$ decreases.","correct":false,"explanation":"A decrease would occur for an exothermic forward reaction. Here, heating favours the endothermic forward reaction, increasing the equilibrium product-to-reactant ratio, so $K_c$ increases."},{"text":"$K_c$ stays constant.","correct":false,"explanation":"Concentration and pressure changes do not alter the equilibrium constant, but temperature changes do. Heating favours the endothermic forward reaction, so $K_c$ increases."},{"text":"$K_c$ becomes infinite.","correct":false,"explanation":"Heating changes the equilibrium position rather than forcing complete conversion into products, so the equilibrium constant does not become infinite. For an endothermic forward reaction, $K_c$ increases."}]},{"id":"24d1475e-e1ba-4a84-a216-b3308f4acbbc","cardType":"cloze","points":null,"front":"$$K_c$ for a given reaction changes only when [ ] changes.","back":"$$K_c$ for a given reaction changes only when temperature changes.","mcqOptions":[{"text":"temperature","correct":true,"explanation":"The equilibrium constant depends on temperature because heating or cooling changes the relative favourability of the forward and reverse reactions. Other changes affect equilibrium position but not $K_c$."},{"text":"pressure","correct":false,"explanation":"Changing pressure can shift an equilibrium involving gases, but the system then re-establishes the same concentration ratio. Only a change in temperature changes $K_c$ for a given reaction."},{"text":"concentration","correct":false,"explanation":"Changing concentration disturbs equilibrium and causes the reaction quotient to return to the existing equilibrium constant. Only a change in temperature changes $K_c$."},{"text":"volume","correct":false,"explanation":"Changing volume may shift a gaseous equilibrium by altering concentrations, but it does not alter the equilibrium constant itself. Only a change in temperature changes $K_c$."}]},{"id":"78cd4c20-f8a2-4fc4-9efc-bdc98cbae60a","cardType":"basic","points":1,"front":"For $aA + bB \\rightleftharpoons cC + dD$, what is the $K_c$ expression?","back":"$K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$","mcqOptions":[{"text":"$K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$","correct":true,"explanation":"Equilibrium concentrations of products appear in the numerator and those of reactants in the denominator. Each concentration is raised to the power of its stoichiometric coefficient."},{"text":"$K_c = \\frac{[A]^a[B]^b}{[C]^c[D]^d}$","correct":false,"explanation":"This reverses the product and reactant terms, giving the reciprocal of the required equilibrium constant. Products belong in the numerator, so the correct expression is $K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$."},{"text":"$K_c = \\frac{[C][D]}{[A][B]}$","correct":false,"explanation":"This omits the powers supplied by the balanced equation’s coefficients. Each equilibrium concentration must be raised to its coefficient, giving $K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$."},{"text":"$K_c = \\frac{c[C]d[D]}{a[A]b[B]}$","correct":false,"explanation":"The coefficients become exponents rather than numerical multipliers of concentration. Applying them as powers gives the correct expression $K_c = \\frac{[C]^c[D]^d}{[A]^a[B]^b}$."}]},{"id":"973c2af3-6e6f-426d-8eed-bab374f81b96","cardType":"cloze","points":null,"front":"In a $K_c$ expression, square brackets mean [ ].","back":"In a $K_c$ expression, square brackets mean equilibrium concentration.","mcqOptions":[{"text":"equilibrium concentration","correct":true,"explanation":"Square brackets denote the molar concentration of each species after equilibrium has been established. These equilibrium concentrations are used to calculate $K_c$."},{"text":"initial concentration","correct":false,"explanation":"Initial concentrations describe the mixture before it reaches equilibrium and cannot generally be substituted directly into the $K_c$ expression. Square brackets represent equilibrium concentration."},{"text":"equilibrium amount","correct":false,"explanation":"Square brackets represent concentration, normally measured in $\\mathrm{mol\\,dm^{-3}}$, rather than the number of moles present. In a $K_c$ expression, this is the equilibrium concentration."},{"text":"equilibrium pressure","correct":false,"explanation":"Partial pressures are used in $K_p$ expressions and are not represented by square brackets. In a $K_c$ expression, square brackets mean equilibrium concentration."}]},{"id":"61367531-fade-445e-9db4-7f8481835c0e","cardType":"cloze","points":null,"front":"A Brønsted-Lowry acid [ ] ions.","back":"A Brønsted-Lowry acid donates H⁺ ions.","mcqOptions":[{"text":"donates H⁺","correct":true,"explanation":"Under the Brønsted-Lowry definition, an acid is a proton donor. It transfers an H⁺ ion to a base during an acid-base reaction."},{"text":"accepts H⁺","correct":false,"explanation":"Accepting H⁺ is the behaviour of a Brønsted-Lowry base, not an acid. A Brønsted-Lowry acid donates H⁺ ions."},{"text":"donates OH^-","correct":false,"explanation":"Donating OH^- relates to hydroxide behaviour and is not the Brønsted-Lowry definition of an acid. A Brønsted-Lowry acid donates H⁺ ions."},{"text":"accepts OH^-","correct":false,"explanation":"Accepting OH^- is not the defining proton-transfer behaviour in Brønsted-Lowry theory. An acid is identified because it donates H⁺ ions."}]},{"id":"b6c51d9b-008c-4aa9-8847-be77112ddff7","cardType":"basic","points":1,"front":"What does a Brønsted-Lowry base do to H⁺ ions?","back":"It accepts H⁺ ions.","mcqOptions":[{"text":"It accepts H⁺ ions.","correct":true,"explanation":"A Brønsted-Lowry base is a proton acceptor. It forms a coordinate bond to an H⁺ ion using an available lone pair of electrons."},{"text":"It donates H⁺ ions.","correct":false,"explanation":"Donating H⁺ is the defining behaviour of a Brønsted-Lowry acid. A Brønsted-Lowry base instead accepts H⁺ ions."},{"text":"It accepts OH^- ions.","correct":false,"explanation":"Accepting OH^- is not the Brønsted-Lowry definition, which is based specifically on proton transfer. A Brønsted-Lowry base accepts H⁺ ions."},{"text":"It donates OH^- ions.","correct":false,"explanation":"Some bases release OH^- in water, but this is the Arrhenius model rather than the general Brønsted-Lowry definition. A Brønsted-Lowry base accepts H⁺ ions."}]},{"id":"1e4f6e3f-858e-4b91-85c1-bbe278954227","cardType":"basic","points":3,"front":"What are the products when an acid reacts with a carbonate?","back":"

Salt
Water
Carbon dioxide

","mcqOptions":[{"text":"

Salt
Water
Carbon dioxide

","correct":true,"explanation":"The acid supplies H⁺ ions, which react with carbonate ions to form water and carbon dioxide. The remaining ions combine to produce a salt."},{"text":"

Salt
Water
Hydrogen

","correct":false,"explanation":"Hydrogen is produced when many metals react with acids, not when carbonates react with acids. A carbonate produces salt, water and carbon dioxide."},{"text":"

Salt
Water
Oxygen

","correct":false,"explanation":"Oxygen is not released in the standard acid-carbonate reaction. Carbonate ions react with H⁺ ions to produce salt, water and carbon dioxide."},{"text":"

Salt
Hydrogen
Oxygen

","correct":false,"explanation":"Hydrogen rather than water is associated with an acid-metal reaction, while oxygen is not produced here. An acid and a carbonate form salt, water and carbon dioxide."}]},{"id":"56f14bd5-a0ee-486c-9316-d0e45210fcf6","cardType":"basic","points":1,"front":"What happens when a weak acid dissolves in water?","back":"It partially dissociates and forms an equilibrium.","mcqOptions":[{"text":"It partially dissociates and forms an equilibrium.","correct":true,"explanation":"A weak acid ionises only to a limited extent in water, so undissociated acid molecules and their ions coexist in a dynamic equilibrium."},{"text":"It completely dissociates and forms no equilibrium.","correct":false,"explanation":"Complete dissociation is characteristic of a strong acid. A weak acid only partially dissociates, establishing an equilibrium between the acid and its ions."},{"text":"It does not dissociate and remains molecular.","correct":false,"explanation":"Weak acids do produce ions in water, but only a fraction of their molecules ionise. They partially dissociate and form a dynamic equilibrium."},{"text":"It completely decomposes and forms new compounds.","correct":false,"explanation":"Dissociation does not mean that a weak acid decomposes irreversibly. It partially forms hydrogen ions and conjugate-base ions in a reversible equilibrium."}]},{"id":"063b47f8-9c3b-4bf8-a36a-b9ddd0d8594f","cardType":"cloze","points":null,"front":"Strong/weak acid strength describes extent of [ ], not concentration.","back":"Strong/weak acid strength describes extent of dissociation, not concentration.","mcqOptions":[{"text":"dissociation","correct":true,"explanation":"Acid strength depends on the proportion of acid molecules that ionise in water. Strong acids dissociate fully, whereas weak acids dissociate only partially."},{"text":"concentration","correct":false,"explanation":"Concentration describes how much acid is present per unit volume, not acid strength. Strength instead describes the extent of dissociation in water."},{"text":"neutralisation","correct":false,"explanation":"Neutralisation describes an acid reacting with a base, rather than the behaviour that defines acid strength. Strength is determined by the extent of dissociation."},{"text":"dilution","correct":false,"explanation":"Dilution lowers an acid solution’s concentration but does not change whether the acid is classified as strong or weak. That classification depends on dissociation."}]},{"id":"9a793077-ab56-4ca6-818b-4032fa109be6","cardType":"basic","points":1,"front":"What is the equation for pH in terms of [H⁺]?","back":"$\\text{pH} = -\\log[\\text{H}^+]$","mcqOptions":[{"text":"$\\text{pH} = -\\log[\\text{H}^+]$","correct":true,"explanation":"By definition, pH is the negative base-10 logarithm of hydrogen-ion concentration. The negative sign makes higher hydrogen-ion concentrations correspond to lower pH values."},{"text":"$\\text{pH} = \\log[\\text{H}^+]$","correct":false,"explanation":"Omitting the negative sign reverses the pH scale and would give negative values for ordinary acidic concentrations. The definition is $\\text{pH}=-\\log[\\text{H}^+]$."},{"text":"$\\text{pH} = -\\ln[\\text{H}^+]$","correct":false,"explanation":"The pH scale uses a base-10 logarithm, not the natural logarithm. The required relationship is $\\text{pH}=-\\log[\\text{H}^+]$."},{"text":"$\\text{pH} = -\\log\\left(\\frac{1}{[\\text{H}^+]}\\right)$","correct":false,"explanation":"Taking the logarithm of the reciprocal incorrectly removes the conventional explicit negative sign. The standard defining equation is $\\text{pH}=-\\log[\\text{H}^+]$."}]},{"id":"2d95323f-b93a-403c-a6f2-72cd5f675c7e","cardType":"basic","points":1,"front":"How is [H⁺] calculated from pH?","back":"$[\\text{H}^+] = 10^{-\\text{pH}}$","mcqOptions":[{"text":"$[\\text{H}^+] = 10^{-\\text{pH}}$","correct":true,"explanation":"Rearranging $\\text{pH}=-\\log[\\text{H}^+]$ requires taking the inverse base-10 logarithm, giving the hydrogen-ion concentration as $10^{-\\text{pH}}$."},{"text":"$[\\text{H}^+] = 10^{\\text{pH}}$","correct":false,"explanation":"The negative sign in the pH definition must remain in the exponent. Using a positive exponent predicts increasing hydrogen-ion concentration as pH increases; the correct expression is $10^{-\\text{pH}}$."},{"text":"$[\\text{H}^+] = \\frac{\\text{pH}}{10}$","correct":false,"explanation":"Dividing pH by 10 is not the inverse of a logarithm. Taking the inverse base-10 logarithm gives $[\\text{H}^+]=10^{-\\text{pH}}$."},{"text":"$[\\text{H}^+] = e^{-\\text{pH}}$","correct":false,"explanation":"Using $e$ incorrectly treats pH as a natural-logarithmic scale. Because pH uses base 10, the concentration is calculated using $[\\text{H}^+]=10^{-\\text{pH}}$."}]},{"id":"9a9d589e-0239-45f3-98d8-0435c6e9e815","cardType":"cloze","points":null,"front":"In $n = cV$, volume $V$ must be measured in [ ].","back":"In $n = cV$, volume $V$ must be measured in dm³.","mcqOptions":[{"text":"dm³","correct":true,"explanation":"When concentration is expressed in mol dm^-3, multiplying by a volume in dm³ cancels the volume units and gives the amount in moles."},{"text":"cm³","correct":false,"explanation":"Using cm³ directly with mol dm^-3 makes the calculated amount 1000 times too large. The volume must first be converted to dm³."},{"text":"m³","correct":false,"explanation":"A volume in m³ is incompatible with concentration in mol dm^-3 unless additional conversion is performed. In $n=cV$, the required unit is dm³."},{"text":"dm³ mol^-1","correct":false,"explanation":"Litres per mole is not a unit of volume alone and cannot be substituted for $V$. With concentration in mol dm^-3, volume must be in dm³."}]},{"id":"d219fd0a-18b4-40b3-a65b-577488ba6c20","cardType":"basic","points":2,"front":"Why repeat the neutralisation without indicator when preparing a pure soluble salt?","back":"To avoid indicator dye contamination.","mcqOptions":[{"text":"To avoid indicator dye contamination.","correct":true,"explanation":"The indicator was needed only to determine the endpoint volume. Repeating the reaction without it prevents coloured indicator molecules from remaining as impurities in the salt."},{"text":"To avoid excess water contamination.","correct":false,"explanation":"An indicator does not significantly dilute the solution, and dilution would not change the number of moles present. The indicator-free repeat prevents dye contamination of the salt."},{"text":"To avoid carbon dioxide contamination.","correct":false,"explanation":"Carbon dioxide contamination is not caused by the indicator in a standard neutralisation. The reaction is repeated without indicator specifically to prevent indicator dye contaminating the product."},{"text":"To avoid excess acid contamination.","correct":false,"explanation":"The repeat uses the accurately measured neutralising volumes, so excess acid should not remain. Omitting the indicator is necessary to prevent its dye contaminating the pure salt."}]},{"id":"ca2b2958-d5b3-4213-bc8e-4bba5cdc0b9e","cardType":"basic","points":1,"front":"In a back titration, what is titrated after adding known excess acid?","back":"The leftover acid.","mcqOptions":[{"text":"The leftover acid.","correct":true,"explanation":"A known excess of acid ensures the original substance reacts completely. Titrating the acid that remains reveals how much acid was consumed by difference."},{"text":"The original reactant.","correct":false,"explanation":"The original reactant is allowed to react with a known excess of acid rather than being titrated directly. The remaining, unreacted acid is what is subsequently titrated."},{"text":"The reaction product.","correct":false,"explanation":"The product is not the species measured in the second titration. A standard solution is used to determine the amount of leftover acid after the initial reaction."},{"text":"The added alkali.","correct":false,"explanation":"The standard alkali is normally delivered from the burette to neutralise the remaining acid; it is not the unknown being determined. The leftover acid is titrated."}]},{"id":"07b34d4e-b329-458e-b2fa-3fa0933c6770","cardType":"basic","points":2,"front":"In NaOH/Na₂CO₃ double titration, what does the extra HCl volume after the phenolphthalein endpoint give?","back":"Moles of Na₂CO₃.","mcqOptions":[{"text":"Moles of Na₂CO₃.","correct":true,"explanation":"After the phenolphthalein endpoint, each carbonate ion needs one further mole of HCl to convert hydrogencarbonate into carbon dioxide and water. The extra HCl therefore equals the carbonate amount."},{"text":"Moles of NaOH.","correct":false,"explanation":"The NaOH has already been fully neutralised by the phenolphthalein endpoint. The additional HCl neutralises hydrogencarbonate derived from carbonate, so its amount gives the original Na₂CO₃ moles."},{"text":"Moles of NaCl.","correct":false,"explanation":"At the phenolphthalein endpoint, carbonate has become hydrogencarbonate, not chloride. The additional HCl amount corresponds to the original Na₂CO₃, rather than to NaCl."},{"text":"Moles of total alkali.","correct":false,"explanation":"The extra HCl is used after all hydroxide has been neutralised, so it cannot measure the total alkali mixture. It specifically equals the moles of Na₂CO₃."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Simple equilibria and acid-base reactions","title":"Simple equilibria and acid-base reactions Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"316f9682-0f4d-59e5-9afa-561eb8789b52","walled":"$L38"}],"$L39","$L3a"]