Calculate the concentration of `H_3O^(+)` of a mixture (solution) that is `0.010M` in `CH_3COOH` and `0.20M` in `NaCH_(3-)COO`. `(K_a=1.8xx10^(-5))`

Explanation

Strategy: Write the suitable equations for both `NaCH_3COO` and `CH_3COOH` and the ionization constant expression for `CH_3COOH`. Then, represent the equilibrium concentrations algebraically and substitute into the `K_a` expression and solve for the unknown.
Solution: The suitable equations and the ionization constant expression are
`Na^(+)CH_3COO^(-)rarr Na^(+)+CH_3COO^(-)` (to completion)
`CH_3COOH+H_2O=H_3O^(+)+CH_3COO^(-)` (reversible)
`K_a=([H_3O^(+)][H_3COO^(-)])/([CH_3COOH])=1.8xx10^(-5)`
This expression of `K_a` is valid for all solutions that contain `CH_3COOH`. In solutions that contain both `CH_3COOH` and `NaCH_3COO`, acetate ions `(CH_3COO^(-))` come from two sources. The ionization constant is satisfied by the total `CH_3COO^(-)` concentration.
Because `NaCH_3COO` is completely dissociated, the `[CH_3COO^(-)]` from `NaCH_3COO` will be `0.20 mol L^(-1)`:
`{:(NaCH_3COO(aq.),rarr,Na^(+)(aq.),+,CH_3COO^(-)(aq.)),(0.20M,,0.0,,0.0),(0.0,,0.20M,,0.20M):}`
Let `x=[CH_3COOH]` that ionizes. Then x is also equal to `[H_3O^(+)]` and `[CH_3COO^(-)]` from `CH_3COOH`. The total concentration of `CH_3COO^(-)` is `(0.20+x)M` and the concentration of unionized `CH_3COOH` is `(0.10-x)M`.
`{:(,CH_(3)COOH(aq.)+H_(2)O(l)hArrH_(3)O^(+)(aq.)+CH_(3)COO^(-)(aq.)),("Initial (M)"," 0.10 0 0.20"),("Change (M)"," -x +x +x"),("Equilibrium (M)",bar(" (0.10-x) x (0.20+x) ")):}`
The substitution of equilibrium concentrations into the ionization constant expression for acetic acid gives
`K_a=(C_(H_3O^+)+C_(CH_3COO^-))/(C_(CH_3COOH))`
`1.8xx10^(-5)=((x)(0.20+x))/((0.10-x))`
The small value of `K_a` suggets that x is very small. This leads to two assumptions:
(i) `x lt lt 0.10`. So `(0.10-x)~~0.10`. This implies that very little `CH_3COOH` ionizes.
(ii) `x lt lt 0.20`. So `(0.20+x)~~0.20`. This implies that most of the `CH_3COO^(-)` comes from `NaCH_3COO` and very little `CH_3COO^(-)` comes from the ionization of `CH_3COOH`.
It is reasonable to assume that `x` (from the ionization of `CH_3COOH`) is small, because `CH_3COOH` is a weak acid, and its ionization is further suppressed by the high concentration of `CH_3COO^(-)` formed by the soluble salt, `NaCH_3COO`.
Applying these assumptions gives
`(0.20x)/(0.10)=1.8xx10^(-5)`
`x=9.0xx10^(-6)`
`:. C_(H_3O^(+))=x=9.0xx10^(-6)`