20 mL of 0.2 M sodium hydroxide solution is added to 40 mL of 0.2 M acetic acid solution. What is the pH of the solution?
`(pK_(a) " of "CH_(3)COOH=4.8)`

Options

A. 9.2
B. 4.8
C. 8.4
D. 2.9

Answer

B

Explanation

Given,
`" "pK_(a) " of "CH_(3)COOH=4.8`
Number of millimoles of NaOH in 20 mL of 0.2 M solution `=20 times 0.2=4 mmol`
Number of millimoles of `CH_(3)COOH` in 40 mL of 0.2 M solution `=40 times 0.2=8mmol`
20 mmol of 0.2 M NaOH will neutralise 20 mmol of 0.2 M acetic acid and produce 4 mmol of sodium acetate as follows :
`CH_(3)COOH+NaOH rarr CH_(3)COOH+H_(2)O`
That is equivalent to 4 mmol of sodium acetate.
Now, `" "pH=pK_(a)+log["Salt"/"Acid"]`
`" "pH=pK_(a)+log[([CH_(3)COONa])/(CH_(3)COOH)]`
`" "pH=4.8+log""([4])/([4])`
`" "pH=4.8`