InterviewSolution
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InterviewSolution
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What would be the pH of a solution obtained by mixing 10 g of acetic acid and 15 g of sodium acetate and making the volume equal to 1L. Dissociation constant of acetic acid at `25^(@)`C is `1.75xx10^(-5)`. |
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Answer» `pH=pK_(a) + log. (["Salt"])/(["Acid"])=pK_(a) + log. ([CH_(3)CO ON a])/([CH_(3)CO OH])` `[CH_(3)CO OH]=(10)/(60) "mol L"^(-1)` (molar mass of `CH_(3)CO OH = 60 g "mol"^(-1)`) `[CH_(3)CO Ona]=(15)/(82) "mol" L^(-1)` (molar mass of `CH_(3)CO Ona = 82 g "mol" ^(-1)`) `pK_(a)=-log K_(a)=-log(1.75xx10^(-5))=5-0.2430 = 4.757` `:. pH = 4.757 + log (15//82)/(10//60) = 4.757+ log. ((15)/(82)xx(60)/(10))` `=4.757 + log 1.098 = 4.757 + 0.0406 = 4.80` |
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