ASAP Show all workout and answer all parts ( Im just reviewing for AP exams so its not graded, but I want it to help me.

Question

A weak acid, HZ, has a Ka value of 6.52 x 10-5.

(a) Write the reaction of HZ with water. 

(b) Write the equilibrium expression for the reaction of HZ with water.

(c) Calculate the pH of a 0.150 M solution of HZ. 

(d) The Y- ion reacts with water to form HY and hydroxide ion, as shown in the following equation.

Y- (aq) + H2O (l)  HY (aq) + OH- (aq)

Given that [OH-] is 1.77 x 10-5 M in a 0.480 M solution of NaY, calculate each of the following:

(i) The value of Kb for the Y- ion. 

(ii) The value of Ka for HY. 

(e) Which acid, HZ or HY, is stronger? Justify your answer.

Answer 1

You've asked this question several times, so I hope that this answer provides what you need to go forward.

a). HZ(aq) + H₂O(l) ==> H₃O⁺(aq) + Z^-(aq)

b). K = [H₃O⁺][Z^-] / [HZ]

c). K = 6.52x10^-5 = (x)(x) / 0.150 - x (assume x is small and ignore it in denominator)

x² = 9.38x10^-6

x = 3.06x10^-3 M = H₃O⁺ (note: this is ~2% of 0.150 M so our assumption above was valid)

pH = -log 3.06x10-3 = 2.51

d). Y^- + H2O ==> HY + OH^-

(i) Kb = [HY][OH^-] / [Y^-] = (1.77x10^-5)(1.77x10^-5) / 0.480 = 6.53x10^-10 = Kb for Y^-

(ii) KaKb = 1x10^-14

Ka = 1x10^-14 / 6.53x10^-10 = 1.53x10^-5 = Ka for HY

e). HZ is a slightly stronger acid than HY because comparing 6.52x10^-5 (for HZ) to 1.53x10^-5 (for HY), HZ will ionize to a greater extent than will HZ given that both are at the same concentration.