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6 gm of CH(3)COOH, 6 gm of NaOH and 6.3 ...

6 gm of `CH_(3)COOH, 6 gm` of `NaOH` and `6.3 gm` of `HNO_(3)` are dissolved in 1L of water. If `K_(a)` of `CH_(3)COOH` is `10^(-5)` then calculate pH of resulting solution.

Text Solution

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The correct Answer is:
5
`.^(n)NaOH = (6)/(40) = 0.15` mole
`.^(n)CH_(3) COOH = (6)/(60) = 0.1` mole
`.^(n)HNO_(3) = (6.3)/(63) = 0.1` mole
First `HNO_(3)` is comletely neutralized. Then
`AcOH + OH^(-) to AcO^(-) + H_(2) O`
0.1 , 0.05
0.05 - 0.05
pH = `pK_(a) = 5`
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