Question

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximately 1. 74 X 10 -5. What is the pH of the resulting solution?

Answer 1

4.76

Step-by-step explanation:

In this case, we have to start with the buffer system:

`CH_3COOH~->~CH_3COO^-~+~H^+`

We have an acid (
`CH_3COOH`) and a base (
`CH_3COO^-`). Therefore we can write the henderson-hasselbach reaction:

`pH~=~pKa+Log([CH_3COO^-])/([CH_3COOH])`

If we want to calculate the pH, we have to calculate the pKa:

`pH=-Log~Ka=4.76`

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

`[CH_3COO^-]=[CH_3COOH]`

If we divide:

`([CH_3COO^-])/([CH_3COOH])~=~1`

If we do the Log of 1:

`Log~1=~zero`

So:

`pH~=~pKa`

With this in mind, the pH is 4.76.

I hope it helps!