# Problem: Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the following solutions:a. 0.100 M HONH2 (Kb = 1.1 x 10-8)b. 0.100 M HONH3Clc. pure H2Od. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl

###### FREE Expert Solution

For a.

0.100 mol

Reaction: HONH2(aq) + HCl(aq) ⇌ HONH3+(aq) + H2O(l)

HONH3+(aq) + H2O(l) ⇌ HONH2(aq) + H3O+(aq)

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}\mathbf{=}{\mathbf{9}}{\mathbf{.}}{\mathbf{09}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{7}}$

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{HONH}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{{{\mathbf{HONH}}_{\mathbf{3}}}^{\mathbf{+}}}}$

${{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\left(x\right)\left(x\right)}{\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\mathbf{x}}$

Solve for x:

${\mathbf{9}}{\mathbf{.}}{\mathbf{09}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{7}}{\mathbf{=}}\frac{{x}^{2}}{\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{-}\overline{)\mathbf{x}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{=}}\sqrt{\left(9.09×{10}^{-7}\right)\left(0.020\right)}$

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###### Problem Details

Calculate the pH after 0.020 mole of HCl is added to 1.00 L of each of the following solutions:

a. 0.100 M HONH2 (Kb = 1.1 x 10-8)
b. 0.100 M HONH3Cl
c. pure H2O
d. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl