pH of a solution?

Question

What is the pH of a solution containing 0.25 M acetic acid and 0.25 M acetate ion. Acetic Acid has Ka= 1.8*10^-5.

Pls,not only answer,clear explanation...thanks.

Answer 1

Use the Henderson-Hasselbach equation to solve this problem.

pH=pKa+log(base/acid)= -log(1.8*10^-5)+log(0.25/0.25)

But since this is a special case where the concentration of the base is equal to the concentration of the acid, then the pH is equal to the equation: pH=pKa (the buffer region).

To get your answer pH= -log(1.8*10^-5)=4.74

Answer 2

Using the Henderson equation,
pH=pKa+log(base/acid)= -log(1.8*10^-5)+log(0.25/0.25) [since its a buffer solution]

your answer pH= -log(1.8*10^-5)

Answer 3

1. Write the Ka equation.

2. List the befores [HAc] = 0.25M, [Ac-]=0.25M. [H+]=0

3. The after equilibrium established [HAc] = 0.25-x, [Ac-]=0.25+x,. H+=x

4. Plug in, Ka=(x)(0.25+x)/(0.25-x)

5. x is small compared to 0.25 (ask your teacher if they haven't explained this yet), so 0.25+x and 0.25-x = 0.25

6. Ka=x(0.25)/0.25 = x

7. x = [H+] = Ka = 1.8x10-5

8. Convert the answer in #7 to pH