The butylammonium ion, C4H9NH3 +, has a Ka of 2.3 × 10^-11.?
C4H9NH3+(aq) + H2O (l) <---> H3O+(aq) + C4H9NH2(aq)
(a) Identify the conjugate base of C4H9NH3+, and calculate Kb for the conjugate base.
(b) Calculate the pH of a 0.025 M solution of the butylammonium ion?
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he conjugate base of the weak acid C4H9NH3+ is C4H9NH2
Kb x Ka = 10 ^14 so Kb = 10 ^-14 / 2.3 × 10^-11 = 4.35 x 10 ^-4
Ka = [H3O+] x [C4H9NH2] / [ C4H9NH3+]
Species HA H3O+ A–
Initial concentration (mol dm-3) ca 10–7 0
Change in concentration (mol dm-3)
–[H3O+] [H3O+] [H3O+]
Equilibrium concentration (mol dm-3)
ca –[H3O+] [H3O+] [H3O+]
Ka = H3O+] * [H3O+] / [ ca –[H3O+]]
the [H3O+] is very small compared to 0.025 so
Ka = [H3O+] ^2 / 0.025
[H3O+] ^2 = 2.3 × 10^-11 X 0.025 = 5.75 x 10 ^-13
[H3O+] = 7.58 x 10 ^-7
pH of the butylammonium acid = 6.12