Acids and Bases Exam Questions and Correct Answers Latest Update 2025 (Rated A+)

Acids and Bases Exam Questions and Correct Answers Latest Update 2025 (Rated A+) Explain why chloroethanoic acid is a stronger acid than ethanoic acid.(2) - Answers Electronegative chlorine withdraws electrons/negative inductive effect Makes O-H bond more polar Explain why data books do not usually contain values of Ka for strong acids.(2) - Answers Strong acids completely dissociate Can't divide by zero in ka as equilibrium lies very far to the right Give the value of the ionic product of water, Kw, measured at 298K, and state its units.(2) - Answers 1x10^-14 mol^2dm^-6 The dissociation of HA into its ions in aqueous solution is an endothermic process. How would its pH change if the temperature were increased? Explain your answer.(3) - Answers Decreases Equilibrium shifts to the right Concentration of H+ increases with temperature Solution B can act as an acidic buffer. Explain what this means and write an equation that shows how Solution B acts as a buffer if a little hydrochloric acid is added - Answers Buffer can resist change in pH on addition of small amounts of H+ or OH- (H+) + (X-) HX Explain the terms acid and conjugate base according to the Brønsted-Lowry theory(2) - Answers Bronsted-Lowry acid is a proton donor It's conjugate base is the substance formed when the acid loses a proton Give the Brønsted-Lowry definition of a base. State the essential feature of an acid-base reaction in aqueous solution, writing an ionic equation to illustrate your answer. - Answers A base is a proton acceptor The essential feature is proton transfer (H+) + (OH-) H2O Explain what is meant by the term weak when applied to acids and bases.(1) - Answers Only partially dissociates in aqueous solution A buffer solution is formed, when approximately half of the original amount of the acid HA(aq) has been neutralised by the base NaOH(aq). Explain how this buffer solution is able to resist change in pH when (i) a small amount of NaOH(aq) is added(2) (ii) a small amount of HCl is added(2) - Answers The added OH- reacts with H+ The equilibrium shifts right to replace lost H_ The added H+ reacts with A- The equilibrium shifts to the left There are two end points that occur when HCl is added to 25cm^3 of sodium carbonate (Na2CO3). Write equations for both these end points(2) - Answers Na2CO3 + HCl HNaCO3 + NaCl HNaCO3 + HCl H2CO3 + NaCl The value of the acid dissociation constant for the monoprotic acid HX is 144 mol dm-3 . What does this suggest about the concentration of undissociated HX in dilute aqueous solution?(1) - Answers Not very big The indicator phenolphthalein is a weak acid which can be represented by the formula HIn. It dissociates in solution and has a pKa value of 9.3. HIn(aq) H+ (aq) + In- (aq) Suggest and explain, with reference to the pKa value, the pH range of phenolphthalein.(2) - Answers at half neutralisation point(says end point on mark scheme?) pH=ka=9.3 pH detectable over 2 pH units (8.3-10.3) State why phenolphthalein is unsuitable for a titration between a strong acid and a weak base.(1) - Answers end point of titration is lower than the phenolphthalein range In a 0.25 M solution, a different acid HY is 95% dissociated. Calculate the value of Ka for the acid HY(6) - Answers [H+] = 0.95 x 0.25 = 0.2375 [HY] = 0.05 x 0.25 = 0.0125 [H+] = [Y-] Ka= [0.2375]^2 / 0.0125 Ka= 4.5125 Define pKa (1) - Answers pKa = -log10Ka Ka = 10^-pKa pKa for HX=3.72. Write an expression for the acid dissociation constant Ka for HX. Use this to show that the pH of any sample of HX is 3.72 when half of the acid has been neutralised by a solution of sodium hydroxide. - Answers Ka=[H+][X-]/[HX] [HX]=[X-] at half neutralisation Hence Ka=[H+] and pKa=pH Explain why indicators cannot be used to determine the end-point of a titration between a weak acid and a weak base.(2) - Answers There is no steep change in pH during a weak acid-base titration Indicator need a sharp rise in pH to change colour quickly By reference to the forces between molecules, explain why ammonia is very soluble in water(2) - Answers Hydrogen bonding Between H2O and NH3 Explain why the pH of a solution containing 1.0 mol dm-3 of ammonia is less than 14 at 298 K.(2) - Answers Ammonia is a weak base Equilibrium to left Write an equation for the reaction of sodium hydroxide with ethanedioic acid (H2C2O4) - Answers H2C2O4 + 2NaOH Na2C2O4 + 2H2O A solution of potassium hydroxide has a pH of 11.90 at 25°C. Calculate the concentration of potassium hydroxide in the solution.(4) - Answers At 25 degrees Celsius (298K) Kw = 10^-14 [H+] = 1.2589x10^-12 [OH-] = 10^-14 / 1.2589x10^-12 = 7.94x10^-3 Explain why dilution with a small volume of water does not affect the pH of a buffer solution.(2)