Answer:
[tex]\large\boxed{\large\boxed{pH=9.39}}[/tex]
Explanation:
1. Chemical equation (given)
[tex]C_5H_5N+H_2O\rightleftharpoons C_5H_5NH^++OH^-[/tex]
2. Chemical equation with the phases:
[tex]C_5H_5N_{(aq)}+H_2O_{(l)}\rightleftharpoons C_5H_5NH^+_{(aq)}+OH^-_{(aq)}[/tex]
3. Equilibrium constant
Only the species in aqueous phase appear in the equilibrium constant.
[tex]K_{eq}=K_b=[C_5H_5NH][OH^-]/[C_5H_5N][/tex]
4. Calculate Kb
[tex]pK_b=-log(K_b)\\ \\ k_b=10^{-pK_b}\\ \\ K_b=10^{-8.75}\\ \\ K_b=1.778\times 10^{-9}[/tex]
5. Write the ICE (initial, change, equilibrium) table for the aqueous species
ICE table:
C₅H₅N C₅H₅NH⁺ +OH⁻
Initial 0.345M 0 0
Change - x + x + x
Equilibrium 0345 - x x x
5. Substitute in the equilibrium constant
[tex]1.778\times 10^{-9}=x^2/(0.345-x)[/tex]
To solve you can neglect x in 0.345 - x, because x is much (very much) less than 0.345M.
[tex]x^2=1.778\times 10^{-9}\times 0.345=6.135\times10^{-10}\\ \\ x=\sqrt{6.135\times10^{-10}}=2.477\times 10^{-5}[/tex]
6. Calculate pOH
7. Calculate pH
