1. a) The pH of a 0.06 M weak monoprotic acid is 3.44. What is the K a for this
ID: 636625 • Letter: 1
Question
1.
a) The pH of a 0.06 M weak monoprotic acid is 3.44. What is the Ka for this weak monoprotic acid?
b) Calculate the pH for a 0.060 M hydrocyanic acid (HCN) solution where the Ka for hydrocyanic acid (HCN) is 4.9 X 10-10.
HCN (aq) <---> H+ (aq) + CN- (aq)
c) Calculate the hydrogen ion concentration in moles per liter (M) for each of the following solutions:
- A solution whose pH is 4.20
- A solution whose hydroxide concentration is 3.7 X 10-9 M
d) Please identify the acid, base, conjugate acid and conjugate base for the following reactions (note A and B are two separate questions):
a) HClO (aq) + CH3NH2 (aq) ? CH3 NH3+ (aq) + ClO- (aq)
b) H2PO4- + NH3 ---><--- HPO42- + NH4+
Explanation / Answer
Ans 1 a) Weak monoprotic acid dissociates into :-
HA ------- H+ + A-
INITIAL 0.060M --- 0 ---- 0 ; X amount of HA dissociates and X amount of H+ and A- is produced.
Change - X ---- +X --- +X
Final 0.060 - X --- X --- X
Ka = [H+][A-]/[HA]
Substitute the values at equillibrium we get,
Ka = [X][X]/[0.06 - X]
Since 0.06 is much much greater than X, we ignore X in the denominator.
Hence, Ka = [X]^2/0.06
[X]= H+ concentration.
pH = - log [H+]
[H+] = 10 ^ - 3.44 = X
Hence, Ka = 10 ^ - 6.88/0.06 Answer.
Answer 1 b ) Ka = [H+][CN-]/[HCN]
At equillibrium Ka = [X]^2/0.06
Hence, on solving we get 4.9 * 10^-10 *0.06 = [X]^2
or, [X] = 0.54 * 0.00001
pH = - log[0.0000054]
Therefore, pH = 5.27 Answer.
Answer 1c) Solution whose pH is 4.2
pH = - log [H+]
[H+] = 10^-4.2 = 1.5 * 10^-4 M
Hence, hydrogen ion concentration is 1.5 * 10^-4.
Given Hydroxide ion concentration = 3.7 * 10^9 M.
pOH = - log [OH-] = - log (3.7 * 10^-9 M)
pH + pOH = 14 ; pH = 14 - {- log (3.7 * 10^ -9 M }
= 14 + log { 3.7 * 10^-9 M}
- log [H+] = 14 + log {3.7 * 10^-9M}
Hence, hydrogen ion concentration H+ = 10^ - { 14 + log(3.7 * 10-9)} M. Answer.
Answer 1d) a) HClO is the acid and its conjugate base is the ClO-.
Ans 1d) b) H2PO4- is the conjugate acid of the base HPO4 2-
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