Question

What mass of benzoic acid, `C_6H_5OOH`, would you dissolve in a total solution volume of 350.0 mL to produce a solution with a pH of 2.85?

`K_a=6.3xx10^-5`

Reaction:
`C_6H_5COOH"(aq)"+H_2O"(l)"\rightleftharpoonsH_3O^(+)"(aq)"+C_6H_5COO^(-)"(aq)"`

Answer 1

I make it approx. `1.3*g`

Explanation:

We address the equilibrium...

`PhCO_2H(aq) + H_2O(l) rightleftharpoonsPhCO_2^(-) + H_3O^+`

And we write the equilibrium expression using the given data...

`K_"eq"=6.30xx10^-5=([H_3O^+][PhCO_2^(-)])/([PhCO_2H])`

But, BY SPECIFICATION, `pH=2.85`...and thus `[H_3O^+]=10^(-2.85)*mol*L^-1=[PhCO_2^(-)]`...and we rearrange the expression to get `[PhCO_2H]`...

`[PhCO_2H]=([H_3O^+][PhCO_2^(-)])/K_a=(10^(-2.85)xx10^(-2.85))/(6.30xx10^-5)`

`-=0.0317*mol*L^-1`

...but this was in a solution of `350*mL`...i.e. a molar quantity with respect to benzoic acid of...

`350*mLxx10^-3*L*mL^-1xx0.0317*mol*L^-1=0.0111*mol`...

...i.e. a mass with respect to benzoic acid of...

`0.0111*molxx122.12*g*mol^-1=1.354*g`...anyway please review my 'rithmetic...