> 0.0240 \\text{ mol}$, the weak acid is in excess."}],"highlights":[{"color":"sky","style":"text-color","anchor":"acid-moles"},{"color":"amber","style":"text-color","anchor":"base-moles"}],"connections":[{"to":"base-moles","from":"acid-moles","color":"sky","label":"compare","style":"solid"}]},"order":2,"title":"Compare moles of reactants","description":"Comparing the initial moles shows that the weak acid is in molar excess."},{"type":"text","order":3,"title":"Explain buffer action","content":"Lactic acid reacts with sodium hydroxide in a $1:1$ molar ratio:\n\n$$\\text{CH}_3\\text{CH(OH)COOH} + \\text{NaOH} \\rightarrow \\text{CH}_3\\text{CH(OH)COONa} + \\text{H}_2\\text{O}$$\n\nBecause the weak acid is in **excess**, the sodium hydroxide is completely neutralised. This reaction leaves behind:\n1. **Unreacted weak acid**: \n $$0.0420\\text{ mol} - 0.0240\\text{ mol} = 0.0180\\text{ mol of } \\text{CH}_3\\text{CH(OH)COOH}$$\n2. **Conjugate base formed**: \n $$0.0240\\text{ mol of } \\text{CH}_3\\text{CH(OH)COO}^- \\text{ (from the sodium salt)}$$\n\nSince the resulting solution contains both a **weak acid** and its **conjugate base**, a buffer solution is formed."}]}],"generatedAt":"2026-06-11T08:03:57.016Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"950b5cde-d184-4b98-a07f-39ea4d068a9b","label":"b","content":"Calculate the pH of the buffer solution that has been prepared. Give your answer to 2 decimal places.","markscheme":"- Calculates the initial amount of lactic acid: $n(\\text{CH}_3\\text{CH(OH)COOH}) = 0.600 \\times \\frac{70.0}{1000} = 0.0420\\text{ mol}$ B1\n- Calculates the initial amount of sodium hydroxide: $n(\\text{NaOH}) = 0.800 \\times \\frac{30.0}{1000} = 0.0240\\text{ mol}$ B1\n- Determines the remaining moles of acid: $0.0420 - 0.0240 = 0.0180\\text{ mol}$, and moles of conjugate base formed: $0.0240\\text{ mol}$ M1\n- Calculates $K_a = 10^{-3.86} = 1.380 \\times 10^{-4}\\text{ mol dm}^{-3}$ and determines $[\\text{H}^+] = 1.380 \\times 10^{-4} \\times \\frac{0.0180}{0.0240} = 1.035 \\times 10^{-4}\\text{ mol dm}^{-3}$ M1\n- Calculates $\\text{pH} = -\\log_{10}(1.035 \\times 10^{-4}) = 3.98$ (to 2 decimal places) A1","marks":5,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"henderson_hasselbalch","label":"Henderson-Hasselbalch Equation","steps":[{"type":"text","order":0,"title":"Calculate initial moles of acid","content":"First, calculate the initial number of moles of lactic acid, $\\text{CH}_3\\text{CH(OH)COOH}$, in the solution using the formula:\n\n$$\\text{Moles} = \\text{Concentration } (\\text{mol dm}^{-3}) \\times \\text{Volume } (\\text{dm}^3)$$\n\nSubstituting the given values:\n\n$$\\begin{aligned} n(\\text{CH}_3\\text{CH(OH)COOH}) &= 0.600 \\text{ mol dm}^{-3} \\times \\frac{70.0}{1000} \\text{ dm}^3 \\\\ &= 0.0420 \\text{ mol} \\end{aligned}$$\n\nThis calculation earns you the first mark.","highlights":[{"text":"70.0 cm^3","location":"specification"},{"text":"0.600 mol dm^{-3}","location":"specification"}]},{"type":"text","order":1,"title":"Calculate initial moles of alkali","content":"Next, calculate the initial moles of the strong base sodium hydroxide, $\\text{NaOH}$:\n\n$$\\begin{aligned} n(\\text{NaOH}) &= 0.800 \\text{ mol dm}^{-3} \\times \\frac{30.0}{1000} \\text{ dm}^3 \\\\ &= 0.0240 \\text{ mol} \\end{aligned}$$\n\nThis calculation earns you the second mark.","highlights":[{"text":"30.0 cm^3","location":"specification"},{"text":"0.800 mol dm^{-3}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$n(\\text{HA})_{\\text{initial}} = \\htmlData{anchor=ha-init}{\\color{@sky}{0.0420}} \\text{ mol}, \\quad n(\\text{OH}^-)_{\\text{initial}} = \\htmlData{anchor=oh-init}{\\color{@amber}{0.0240}} \\text{ mol}$"},{"latex":"$$n(\\text{HA})_{\\text{remaining}} = \\htmlData{anchor=ha-init-sub}{\\color{@sky}{0.0420}} - \\htmlData{anchor=oh-sub-1}{\\color{@amber}{0.0240}} = \\htmlData{anchor=ha-rem}{\\color{@purple}{0.0180}} \\text{

Acids, bases and buffers

Question 23

Acids, bases and buffers Questions

Acids, bases and buffers

Question 23

Acids, bases and buffers Questions