Question

Enter your answer in the provided box. Calculate the pH at the equivalence point for the following titration: 0.40 M HCl versus 0.40 M methylamine (CH3NH2). The Ka of methylammonium is 2.3 × 10−11

Answer #1

CH_{3}NH_{2} + HCl --->
CH_{3}NH_{3}^{+} + Cl^{-}

Suppose 1 L of CH_{3}NH_{2}

since the concentrations of HCl and CH_{3}NH_{2}
are the same we need 1 L of HCl to reach the equivalence
point

Total volume = 2

At the equilvalence point 0.40 mol CH_{3}NH_{2} +
0.40 mole H+ ----> 0.40mole
CH_{3}NH_{3}^{+}

[CH_{3}NH_{3}^{+}] = 0.40 / 2 L = 0.20
M

CH_{3}NH_{3}^{+} + H_{2}O
<-----> CH_{3}NH_{2} +
H_{3}O^{+}

Ka = 2.3 x 10^{-11}

Ka = [H_{3}O^{+}][CH_{3}NH_{2}]
/ [CH_{3}NH_{3}^{+}]

at equilibrium [CH_{3}NH_{2}] =
[H_{3}O^{+}] = x,
[CH_{3}NH_{3}^{+}] = 0.20 - x

Therefore

2.3 x 10^{-11} = x^{2} / 0.20 - x

since 0.20 >>> x

2.3 x 10^{-11} = x^{2} / 0.20

x = [H+]= 2.14 x 10^{-6} M

pH = - log (2.14 x 10^{-6})

**pH = 5.66**

Calculate the pH at the equivalence point for the following
titration: 0.40 M HCl versus 0.40 M methylamine (CH3NH2). The Ka of
methylammonium is 2.3 × 10^−11

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