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The pH of a solution containing 0.1 mol ...

The pH of a solution containing 0.1 mol of `CH_(3)COOH, 0.05` mol of `NaOH, and 0.2 mol ofCH_(3)COONa,` in 1.L. `(pK_(a) of CH_(3)COOH=4.74)` is :

A

`5.44`

B

`5.20`

C

`5.04`

D

`4.74`

Text Solution

Verified by Experts

The correct Answer is:
A
`CH _(3) COOH + Na OH to CH_(3) COONa+ H _(2) O`
`{:("Initial mol" , 0.1 , 0.05, 0, 0),("Final ol", (0.1 0.05), (0.05 -0.05), 0.05, -), (, =0.05, =0,,):}`
Total moles of `CH _(3) COONa = 0.2 + 0.05 = 0. 25 therefore [CH (3) COONa] = (0.25 mol )/(1L) = 0. 25 M`
Moles of `CH _(3) COOH ` left `= (0.1 - 0.05) = 0.05 therefore [CH _(3) COOH] = (0.05mol)/(1L) = 0.05M`
Thus , acidic buffer is formed
`pH = pK _(a) + log ""(["Salt"])/(["Acid"])= 4.74 + log ((0. 25)/( 0. 0 5)) = 4. 74 + log 5 = 4.74 + 0. 7 = 5.44`
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