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**Problem**: In the titration of 50.0 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 x 10-4), with 0.50 M HCl, calculate the pH under the following conditions.a. after 50.0 mL of 0.50 M HCl has been added

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(1) Calculated moles of w.base and s.acid

$\mathrm{mol}{\mathrm{CH}}_{3}{\mathrm{NH}}_{2}=\left(50\overline{)\mathrm{mL}}\right)\left(\frac{1\mathrm{x}{10}^{-3}\overline{)\mathrm{L}}}{1\overline{)\mathrm{mL}}}\right)\left(\frac{1\mathrm{mol}}{1\overline{)\mathrm{L}}}\right)$

mol CH_{3}NH_{2} = 0.05 mol

$\mathrm{mol}\mathrm{HCl}=\left(50\overline{)\mathrm{mL}}\right)\left(\frac{1\mathrm{x}{10}^{-3}\overline{)\mathrm{L}}}{1\overline{)\mathrm{mL}}}\right)\left(\frac{0.5\mathrm{mol}}{1\overline{)\mathrm{L}}}\right)$

mol HCl = 0.025 mol

(2) Construct an ICF chart:

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In the titration of 50.0 mL of 1.0 M methylamine, CH_{3}NH_{2} (K_{b} = 4.4 x 10^{-4}), with 0.50 M HCl, calculate the pH under the following conditions.

a. after 50.0 mL of 0.50 M HCl has been added

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