Answer:
8.6
Explanation:
Using the expression :
[tex]K_a\times K_b=K_w[/tex]
Where, [tex]K_w[/tex] is the dissociation constant of water.
At [tex]25\ ^0C[/tex], [tex]K_w=10^{-14}[/tex]
Thus, for benzoic acid , pKa = 4.20
Thus, [tex]K_a=10^{-4.20}=6.31\times 10^{-5}[/tex]
[tex]K_b[/tex] for Sodium benzoate can be calculated as:
[tex]K_a\times K_b=K_w[/tex]
[tex]6.31\times 10^{-5}\times K_b=10^{-14}[/tex]
[tex]K_b=1.58\times 10^{-10}[/tex]
The benzoate ion will dissociate as:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
[tex]K_b[/tex] expression is:-
[tex]K_{b}=\frac {\left [ C_6H_5COOH^{+} \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Given that:-
Moles = 0.100 moles
Volume = 1.00 L
Thus, Concentration = 0.100/ 1.00 M = 0.1 M
Considering the ICE table in the image below.
So,
[tex]1.58\times 10^{-10}=\frac{x^2}{0.1-x}[/tex]
[tex]1.58\left(0.1-x\right)=10^{10}x^2[/tex]
Solving for x, we get that, [tex]x=3.97\times 10^{-6}[/tex]
Thus, [tex][OH^-]=3.97\times 10^{-6}[/tex]
Also,
[tex]pOH=-log[OH^-]=-log(3.97\times 10^{-6})=5.4[/tex]
Also, pH + pOH = 14
So, pH = 14 - 5.4 = 8.6
