Answer:
Explanation:
Since [tex]CH _ 3COOH[/tex] is a weak acid to find the pH of [tex]CH _ 3COOH[/tex] we use the formula
[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]
where
Ka is the acid dissociation constant
c is the concentration
From the question
Ka of [tex]CH _ 3COOH[/tex] = 1.75 × 10^-5
c = 1.00 × 10-³M
Substitute the values into the above formula and solve for the pH
That's
[tex]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 \: \: \: \: \: \: \: \: [/tex]
We have the answer as
To find the pOH we use the formula
pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 3.9
We have the answer as
Hope this helps you
