Question

Determine the pH of each solution.

a. 0.155 M HNO2 (for HNO2, Ka=4.6×10^−4)

b. 0.0210 M KOH

c. 0.240 M CH3NH3I (for CH3NH2, Kb=4.4×10^−4)

d. 0.324 M KC6H5O (for HC6H5O, Ka=1.3×10^−10)

Answer 1

a)
for simplicity lets write weak acid as HA

HA          ----->     H+   +   A-
0.1550                 0         0
0.1550-x               x         x

Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((4.6E-4)*0.1550) = 8.44E-3

pH = -log [H+] = -log (8.44E-3) = 2.07

Answer: 2.07

b)
[OH-] = [KOH] = 0.0210 M = 2.1E-2 M

pOH = -log [OH-] = -log (2.10E-2) = 1.68

PH = 14 - pOH = 14 - 1.68 = 12.32

Answer: 12.32

c)
Here weak base is NH2-
for simplicity lets write weak base as A-

A-        + H2O   ----->     AH   +   OH-
0.2400                        0         0
0.2400-x                      x         x

Kb = [AH][OH-]/[B-]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((4.4E-4)*0.2400) = 1.03E-2

pOH = -log [OH-] = -log (1.03E-2) = 1.99

PH = 14 - pOH = 14 - 1.99 = 12.01
Answer: 12.01

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