Question

What is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?

Answer 1

Answer:

pH = 3.9

pOH = 10.1

Explanation:

Since $CH _ 3COOH$ is a weak acid to find the pH of $CH _ 3COOH$ we use the formula

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

where

Ka is the acid dissociation constant

c is the concentration

From the question

Ka of $CH _ 3COOH$ = 1.75 × 10^-5

c = 1.00 × 10-³M

Substitute the values into the above formula and solve for the pH

That's

$pH = \frac{1}{2} ( - log(1.75 \times {10 }^{ - 5} - log(1.00 \times {10}^{ - 3} ) ) \\ = \frac{1}{2} (4.757 + 3) \\ = \frac{1}{2} \times 7.757) \\ = 3.8785 \: \: \: \: \: \: \: \:$

We have the answer as

pH = 3.9

To find the pOH we use the formula

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.9

We have the answer as

pOH = 10.1

Hope this helps you