# What is the pH of a 0.470 M solution of methylamine?

May 6, 2017

$p H = 12$

#### Explanation:

We need (i) $p {K}_{B}$ values for methylamine, alternatively $p {K}_{a}$ values for ${H}_{3} C N {H}_{3}^{+}$. This site gives $p {K}_{b} = 3.66$.

And (ii) we need a stoichiometric equation.........

${H}_{3} C N {H}_{2} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} C N {H}_{3}^{+} + H {O}^{-}$

And by definition, this equilibrium is governed by the quotient.....

${K}_{b} = {10}^{- p {K}_{b}} = {10}^{- 3.66} = \frac{\left[{H}_{3} C N {H}_{3}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{3} C N {H}_{2}\right]}$.

Now if initially, $\left[{H}_{3} C N {H}_{2}\right] = 0.470 \cdot m o l \cdot {L}^{-} 1$, and we say the amount of dissociation is $x$, then we can write:

${K}_{b} = \frac{\left[{H}_{3} C N {H}_{3}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{3} C N {H}_{2}\right]} = \frac{\left(x\right) \times \left(x\right)}{0.470 - x} = {x}^{2} / \left(0.470 - x\right) = {10}^{- 3.66}$.

This is a quadratic in $x$, which we could solve exactly, but because chemist are lazy folk, we make the approximation, that $0.366 \text{>>} x$, and that ${x}^{2} / \left(0.470 - x\right) = {10}^{- 3.66} \cong {x}^{2} / \left(0.470\right)$.

And thus ${x}_{1} = \sqrt{{10}^{- 3.66} \times 0.470} = 1.01 \times {10}^{-} 2$, and if we recycle this first approximation back into the equation, we gets.....

${x}_{2} = \sqrt{{10}^{- 3.66} \times \left(0.470 - 1.01 \times {10}^{-} 2\right)} = 1.00 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

${x}_{3} = \sqrt{{10}^{- 3.66} \times \left(0.470 - 1.00 \times {10}^{-} 2\right)} = 1.00 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$

Because the approximations have converged, we are willing to accept this value. But $x = \left[H {O}^{-}\right]$ by definition; and so $p O H = - {\log}_{10} \left(1.00 \times {10}^{-} 2\right) = + 2$

And since we know (or should know) that $p O H + p H = 14$, $p H = 12$.