Question

At 25°C, what is the hydronium ion concentration in:

a. 0.350 M sodium chloroacetate (Ka=1.36e-3)?

b. 4.5e-3 aniline hydrochloride (Kb=3.98e-10)?

c. 0.520 M HIO3 (Ka=1.7e-1)?

Answer 1

a.) 0.350 M sodium chloroacetate (Ka=1.36e-3)

ClCH2COO-   + H2O -----------------------> ClCH2COOH + OH-

0.350                                                           0                         0 ------------> intial

0.350-x                                                       x                          x -----------------> equilibrium

Kb = Kw/ Ka = 1.0 x 10^-14 / 1.3 x 10^-3 = 7.35 x 10^-12

Kb = x^2 / 0.350-x

7.35 x 10^-12 = x^2 / 0.350-x

x^2 + 7.35 x 10^-12x - 2.57 x 10^-12 = 0

x = 1.60 x 10^-6

x = [OH-] = 1.60 x 10^-6 M

[H3O+ ] = hydronium ion = 1.0 x 10^-14 / 1.60 x 10^-6

[H3O+ ] = hydronium ion = 6.24 x 10^-9 M

b. 4.5e-3 aniline hydrochloride (Kb=3.98e-10)?

aniline hydrochloride is the salt of weak base and strong base .so pH <7

pH = 7 - 1/2 [pKb +logC]

pH = 7 - 1/2 [9.40 + log 4.5 x 10^-3]

pH = 3.47

[H3O+] = 10^-pH

[H3O+]    = 3.39 x 10^-4 M

c. 0.520 M HIO3 (Ka=1.7e-1)?

[H3O+] = 0.224 M