Question

Determine the pH of each of the following two-component solutions.

1. 0.130 M in NH₄Cl and 0.155 M in HNO₃

2. 8.86

Answer 1

1)

HN03 is a very strong acid and NH4Cl is a very weak acid

so

the contribution of H+ is predominantly from HN03

we know that

Hn03 ----> H+ + N03-

from the above reaction

[H+] = [HN03] = 0.155 M

now

pH = -log [H+]

pH = -log 0.155

pH = 0.809

2)

KBr is a salt of strong acid and strong base

so it is nuetral and does not react . so K+ and Br- are spectator ions here

now

sodium benzoate is a weak base with Kb = 1.56 x 10-10

for weak bases

[OH-] = sqrt ( Kb x C)

[OH-] = sqrt ( 1.56 x 10-10 x 8.86 x 10-2 )

[OH-] = 3.72 x 10-6

now

pOH = -log [OH-]

pOH = -log 3.72 x 10-6

pOH = 5.43

now

pH = 14 - pOH

pH = 14 - 5.43

pH = 8.57

3)

HCl and HN03 are both strong acids

HCL ----> H+ + Cl-

[H+] = [HCl]

similarly

[H+] = [HN03]

[H+] total = [H+] from HCl + [H+} from HN03

[H+] total = 4 x 10-2 + 2.3 x 10-2

[H+] total = 6.3 x 10-2

pH = -log [H+]

pH = -log 6.3 x 10-2

pH = 1.2

so the pH is 1.2