Question

A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coefficient of H+.B. The measured pH of a solution of 0.010 HCl and 0.090 KCl at 25 degree Celsius is 2.102. Calculate the activity coefficient of H+ in this solution.C. Why does the pH change in part B relative to that in part A?

Answer 1

Answer:

activity coefficient $\mathbf{\gamma =0.809}$

activity coefficient $\mathbf{\gamma = 0.791}$

The change in pH in part A = 0.092

The change in pH in part B =  0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = $10^{-1.092}$

[a] = 0.0809 M

Recall that [a] = $\gamma$ × C

where;

$\gamma$  = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

$\gamma = \dfrac{[a]}{C}$

$\gamma = \dfrac{0.0809 \ M}{0.100 \ M}$

$\mathbf{\gamma =0.809}$

B.

The pH of a solution of HCl and KCl = 2.102

[a] = $10^{-2.102}$

[a] = 0.00791 M

activity coefficient $\gamma = \dfrac{0.00791 \ M}{0.01 \ M}$

$\mathbf{\gamma = 0.791}$

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102