Hi. Someone answered these but 1. The salt HONH,Cl is a strong electrolyte whose
ID: 1025569 • Letter: H
Question
Hi. Someone answered these but 1. The salt HONH,Cl is a strong electrolyte whose cation behaves as a weak acid. (aq) H* (aq) HON, (aq) + HONH3 + a. Compute the K. associated with this cation. b. Use an ICB table to compute the H concentration and pH of a solution containing 0.664 M HONH,Ci. 2. The salt KCN is a strong electrolyte whose anion behaves as a weak base. CN- (aq) + H2O HCN (aq) + OH- (aq) a. Compute the K associated with this anion. b. Use an ICE table to compute the OH- concentration and pH of a solution containing 0.306 M KCNExplanation / Answer
1. Kb for NH2OH (literature) = 1.10 x 10^-8
a. So Ka for HONH3+ = Kw/Kb
= 1 x 10^-14/1.10 x 10^-8
= 9.1 x 10^-7
b. [HONH3Cl] = 0.664 M
or,
[HONH3+] = 0.664 M
ICE chart
HONH3+ <===> HONH2 + H+
I 0.664 - -
C -x +x +x
E 0.664-x x x
So,
Ka = [HONH2][H+]/[HONH3+]
9.1 x 10^-7 = x^2/(0.664-x)
x^2 + 9.1 x 10^-7x - 6.04 x 10^-7 = 0
x = [H+] = 7.77 x 10^-4 M
pH = -log[H+]
= -log(7.77 x 10^-4) = 3.11
==
2. For CN-
Ka for HCN (literature) = 4.9 x 10^-10
a. Kb for CN- = Kw/Ka
= 1 x 10^-14/4.9 x 10^-10
= 2.04 x 10^-5
b. [KCN] = 0.306 M
or,
[CN-] = 0.306 M
ICE chart
CN- + H2O <===> HCN + OH-
I 0.306 - - -
C -x - +x +x
E 0.306-x - x x
So,
Kb = [HCN]OH-]/[CN-]
2.04 x 10^-5 = x^2/(0.306-x)
x^2 + 2.04 x 10^-5x - 6.24 x 10^-6 = 0
x = [OH-] = 2.50 x 10^-3 M
pOH = -log[OH-] = 2.60
pH = 14 - pOH
= 14 - 2.60
= 11.4
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