# Question 0a0b2

Apr 13, 2015

!! LONG ANSWER !!

After the addition of the hydrochloric acid, the pH of your buffer will be 10.79.

So, you're dealing with a buffer solution comprised of dimethylamine, ${\left(C {H}_{3}\right)}_{2} N H$, a weak base, and dimethylammonium chloride, ${\left(C {H}_{3}\right)}_{2} N {H}_{2} C l$, its conjugate acid.

Before doing any calculations, try to predict what's going to happen to the pH of the solution once hydrochloric acid, a strong acid, is added.

The weak base will react with the acid to produce the base's conjugate acid; as a result of that reaction, the hydrochloric acid will be consumed.

At the same time, the number of moles of dimethylamine will decrease, and the number of moles of dimethylammonium ions will increase.

The balanced chemical equation for this reaction is (all species are in aqueous solution, so I won't write the (aq) subscripts)

${\left(C {H}_{3}\right)}_{2} N H + H C l \to {\left(C {H}_{3}\right)}_{2} N {H}_{2} C l$

The net ionic equation is

${\left(C {H}_{3}\right)}_{2} N H + {H}^{+} \to {\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+}$

So, start by calculating how many moles of dimethylamine and dimethylammonium chloride you initially have in solution

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{a \min e} = C \cdot V = \text{0.30 M" * "1.0 L" = "0.30 moles}$

${n}_{a m m o n i u m} = \text{0.25 M" * "1.0 L" = "0.25 moles}$

Now you add the $H C l$, more specifically you add

0.73cancel("g") * "1 mole"/(36.5cancel("g")) = "0.02 moles HCl"#

Use the net ionic equation to set up an ICE table

$\text{ } {\left(C {H}_{3}\right)}_{2} N H + {H}^{+} \to {\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+}$
I.....0.30....................0.02..............0.25
C....(-0.02)...............(-0.02)............(+0.02)
E.....0.28.....................0..................0.27

The new concentrations of the weak base and its conjugate acid will be

$\left[{\left(C {H}_{3}\right)}_{2} N H\right] = \text{0.28 moles"/"1.0 L" = "0.28 M}$

$\left[{\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+}\right] = \text{0.27 moles"/"1.0 L" = "0.27 M}$

Now use the Henderson-Hasselbalch equation to solve for pOH

$p O H = p {K}_{b} + \log \left(\frac{\left[{\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+}\right]}{\left[{\left(C {H}_{3}\right)}_{2} N H\right]}\right)$

$p O H = 3.23 + \log \left(\text{0.27 M"/"0.28 M}\right) = 3.23 - 0.016 = 3.21$

Therefore, the pH will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 3.21 = \textcolor{g r e e n}{10.79}$

SIDE NOTE The initial pH of the solution is

$p O H = 3.23 + \log \left(\text{0.25 M"/"0.30 M}\right) = 3.15$

$p {H}_{\text{initial}} = 14 - 3.15 = 10.85$

Notice that the addition of a strong acid reduced the pH of the buffer only slightly.