When HNO2\text{HNO}_2HNO2 is added to water at 298 K, this equilibrium is established.
HNO2(aq)⇌H+(aq)+NO2−(aq) \text{HNO}_2\text{(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{NO}_2^-\text{(aq)} HNO2(aq)⇌H+(aq)+NO2−(aq)At equilibrium, [HNO2]=5.50×10−2 mol dm−3[\text{HNO}_2] = 5.50 \times 10^{-2}\text{ mol dm}^{-3}[HNO2]=5.50×10−2 mol dm−3 and [NO2−]=4.80×10−3 mol dm−3[\text{NO}_2^-] = 4.80 \times 10^{-3}\text{ mol dm}^{-3}[NO2−]=4.80×10−3 mol dm−3. What is the value of the equilibrium constant, in mol dm−3\text{mol dm}^{-3}mol dm−3, at 298 K?
8.73×10−28.73 \times 10^{-2}8.73×10−2
1.15×1011.15 \times 10^{1}1.15×101
2.30×10−52.30 \times 10^{-5}2.30×10−5
4.19×10−44.19 \times 10^{-4}4.19×10−4
77 exam-style questions on AQA A Level Chemistry Acids and bases (A-level only), covering Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only), Definition and determination of pH (A-level only), The ionic product of water, $K_w$ (A-level only), Weak acids and bases $K_a$ for weak acids (A-level only), pH curves, titrations and indicators (A-level only), and Buffer action (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.