Question

Approximately how much water should be added to 10.0mL of 11.1M HCl so that it has the same pH as 0.90M acetic acid (K_a=1.8 × 10^-5)?

a) 28mL

b) 276mL

c) 3L

d) 28 L

e) 276 L

Answer 1

Step 1: Calculate [H+] from CH3COOH
CH3COOH dissociates as:

CH3COOH          ----->     H+   + CH3COO-
0.9                 0         0
0.9-x               x         x


Ka = [H+][CH3COO-]/[CH3COOH]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.8*10^-5)*0.9) = 4.025*10^-3

since c is much greater than x, our assumption is correct
so, x = 4.025*10^-3 M



So, [H+] = x = 4.025*10^-3 M

Step 2:
For same pH, [H+] should be same.
So, [H+] from HCl = 4.025*10^-3 M

Since HCl is strong acid,
[HCl] = [H+] = 4.025*10^-3 M

So, we need to dilute HCl so that its final concentration is 4.025*10^-3 M

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 11.1 M
M2 = 4.025*10^-3 M
V1 = 10 mL

use:
M1*V1 = M2*V2
V2 = (M1 * V1) / M2
V2 = (11.1*10)/(4.025*10^-3)
V2 = 27578 mL

So,
Volume of H2O added = V2 - V1
= 27578 mL - 10.0 mL
= 27568 mL
= 27.6 L
Answer: d) 28 L