A 50-mL sample of a solution containing an unknown weak base was titrated with 0

Question

A 50-mL sample of a solution containing an unknown weak base was titrated with 0.1000 M HCl. The titration curve is shown below. a. Explain how you would use the titration curve to determine the value of Kb for the unknown weak base. b. Explain why the solution is acidic at the equivalence point in the titration. Support your argument(s) with appropriate chemical equations. Consider the following aqueous salt solutions, I. 0.10 M NaNO2 II. 0.10 M NaF III. 0.10 M HCOONa Which of these solutions would be expected to have the highest pH, and which solution would be expected to have the lowest pH? Explain your reasoning. (HNO2, Ka = 7.1 x 10^-4; HF, Ka = 6.8 x 10^-4; HCOOH, Ka = 1.8 x 10^-4)

Explanation / Answer

a) From the titration curve, determine the equivalence point, which is when 50mL of HCl was added.

pKa is the pH at half-equivalence point, that is pKa is the pH of the mixture when 25mL of HCl was added, which is around 6 in the problem shown above.

pKb - 14 - pKa = 14-6 = 8. Therefore, Kb = 1x10-8

b) B + HCl ------> BH+ + Cl- + H2O

At equivalence point, all of the base are protonated. Hence in order to maintain the equilibrium, some of the protonated base releases protons and forms base. The protons decreases the pH and hence the solution is acidic.

c) let x be the amount of acid formed when the compounds dissolved in water.   HA + OH- <-------> A- + H2O

Ka = [A-]/[HA][OH-] = [A-][H+]/[HA]Kw => [H+] = Ka*Kw*[HA]/[A-]

So a higher Ka results in higher higher [H+].

Therefore, the lowest pH would be formed by sodium formate and highest pH would be formed by NaNO2 solution.

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