Question

(1) A 0.10 M HF solution is 8.4% ionized. Calculate the H+ ion concentration.

(2) Determine the pH of a KOH solution made by mixing 0.251 g KOH with enough water to make 1.0 × 102 mL of solution

Please answer both clearly. I am stuck on these 2 questions. Thank you.

Answer 1

1)
HF   <->   H+   +   F-
0.10       0       0   (initial)
0.10-x   x       x   (at equilibrium)

% ionisation = x*100/c
8.4 = x*100/0.10
x = 8.4*10^-3 M

So,
[H+] = 8.4*10^-3 M

Answer: 8.4*10^-3 M

2)

Molar mass of KOH,
MM = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol


mass(KOH)= 0.251 g

use:
number of mol of KOH,
n = mass of KOH/molar mass of KOH
=(0.251 g)/(56.11 g/mol)
= 4.474*10^-3 mol
volume , V = 1*10^2 mL
= 0.1 L


use:
Molarity,
M = number of mol / volume in L
= 4.474*10^-3/0.1
= 4.474*10^-2 M

So,
[OH-] = 4.474*10^-2 M

Given:
[OH-] = 4.474*10^-2 M

use:
pOH = -log [OH-]
= -log (4.474*10^-2)
= 1.3493


use:
PH = 14 - pOH
= 14 - 1.3493
= 12.6507
Answer: 12.65