pH of a mixture which is 0. 1 M in `CH_(3)COOH` and 0.05 M in `(CH_(3)COOH)_(2)Ba` is `[pK_(a)` of `CH_(3)COOH=4.74]`
pH of a mixture which is 0. 1 M in `CH_(3)COOH` and 0.05 M in `(CH_(3)COOH)_(2)Ba` is `[pK_(a)` of `CH_(3)COOH=4.74]`
A. 4.74
B. 5.04
C. 4.44
D. `7.00`
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Correct Answer - A
`[CH_(3)COO^(-)]=2[(CH_(3)COO)_(2)Ba]`
`=2 xx 0.05 =0.1 M`
`[CH_(3)COOH]=0.1M`
`pH=pK_(a)+log.(["Salt"])/(["Acid"])`
`=4.74 +log.(0.1)/(0.1)=4.74`
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