Question

The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.

Answer 1

Answer:

The answer is

$Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3}$

Explanation:

The Ka of an acid when given the pH and concentration can be found by

$pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c)$

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

$5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010)$

$- \frac{1}{2} log(Ka) = 5.82 + 1$

$- \frac{1}{2} log(Ka) = 6.82$

Multiply through by - 2

$log(Ka) = - 13.64$

Find antilog of both sides

We have the final answer as

$Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3}$

Hope this helps you