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vladimir2022 [97]
3 years ago
9

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

Chemistry
1 answer:
blagie [28]3 years ago
8 0

Answer:

The answer is

<h3>Ka = 2.29 \times  {10}^{ - 14} moldm^{ - 3}</h3>

Explanation:

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

<h3>pH =  -  \frac{1}{2}  log(Ka)  -  \frac{1}{2}  log(c)</h3>

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

<h3>5.82 =   - \frac{1}{2}  log(Ka)  -  \frac{1}{2}  log(0.010)</h3><h3 /><h3>-  \frac{1}{2}  log(Ka)  = 5.82 + 1</h3><h3 /><h3>-  \frac{1}{2}  log(Ka)  = 6.82</h3>

Multiply through by - 2

<h3>log(Ka)  =  - 13.64</h3>

Find antilog of both sides

We have the final answer as

<h3>Ka = 2.29 \times  {10}^{ - 14} moldm^{ - 3}</h3>

Hope this helps you

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