# Question #eed26

Apr 26, 2015

The pH of the solution will be 4.77.

So, you know that you're dealing with citric acid, a triprotic acid that ionizes in three steps, each with its own acid dissociation constant - ${K}_{a 1}$, ${K}_{a 2}$, and ${K}_{a 3}$, respectively.

${H}_{3} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{2} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$, ${K}_{a 1}$

${H}_{2} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{2 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$, ${K}_{a 2}$

$H {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{3 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$, ${K}_{a 3}$

The key to this problem is actually the relationship that exists between how much citric acid you have, 0.001 moles, and how many moles of strong base you add.

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

${n}_{N a O H} = \text{0.1 M" * 15 * 10^(-3)"L" = "0.0015 moles NaOH}$

The balanced chemical equation for the neutralization of citric acid is

${H}_{3} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)} + 3 N a O {H}_{\left(a q\right)} \to N {a}_{3} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)} + 3 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : 3$ mole ratio between citric acid and sodium hydroxide. This means that a complete neutralization would required 3 times more moles of sodium hydroxide than the number of moles of citric acid you start with.

Since you have less moles of sodium hydroxide than you would need for a complete neutralization, you can predict that the pH will be acidic, i.e. some citric acid will be left in solution.

So, start by writing the first reaction between citric acid and sodium hydroxide (net ionic equation, so I didn't include the sodium cation)

${H}_{3} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)} + O {H}_{\left(a q\right)}^{-} \to {H}_{2} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

Once again, notice the $1 : 1$ mole ratio that exists between all the species involved. When you react 0.001 moles of citric acid with 0.0015 moles of sodium hydroxide, all of the ${H}_{3} {C}_{6} {H}_{5} {O}_{7}$ will be consumed.

As a result, you'll produce some sodium citrate, $N a {H}_{2} {C}_{6} {H}_{5} {O}_{7}^{-}$.

${n}_{{H}_{3} {C}_{6} {H}_{5} {O}_{7}} = 0.001 - 0.001 = 0$

${n}_{O {H}^{-}} = 0.0015 - 0.001 = \text{0.0005 moles}$

${n}_{{H}_{2} {C}_{6} {H}_{5} {O}_{7}^{-}} = \text{0.001 moles}$

Since you still have strong base left in solution, a second reaction will take place

${H}_{2} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{-} + O {H}_{\left(a q\right)}^{-} \to H {C}_{6} {H}_{5} {O}_{7 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)}$

This time, the strong base will be consumed, and disodium citrate will be produced, so you'll be left with

${n}_{{H}_{2} {C}_{6} {H}_{5} {O}_{7}^{-}} = 0.001 - 0.0005 = \text{0.0005 moles}$

${n}_{H {C}_{6} {H}_{5} {O}_{7}^{2 -}} = 0 + 0.0005 = \text{0.0005 moles}$

At this point, this should look familiar. Equal number of moles, i.e. concentrations, since the volume is the same for both, for a weak acid, ${H}_{2} {C}_{6} {H}_{5} {O}_{7}^{-}$, and its conjugate base, $H {C}_{6} {H}_{5} {O}_{7}^{2 -}$.

Use the Henderson-Hasselbalch equation to calculate the pH

$p {H}_{\text{sol}} = p {K}_{a 2} + \log \left(\frac{\left[H {C}_{6} {H}_{5} {O}_{7}^{2 -}\right]}{\left[{H}_{2} {C}_{6} {H}_{5} {O}_{7}^{-}\right]}\right)$

$p {H}_{\text{sol}} = - \log \left({K}_{a 2}\right) + \log \left(1\right) = p {K}_{a 2} = \textcolor{g r e e n}{4.77}$

Apr 26, 2015

$\text{pH} = 4.76$

Since citric acid is tri - protic I'll call it ${H}_{3} X$.

Sodium hydroxide is a strong base so will strip all the protons from the citric acid:

${H}_{3} X + 3 N a O H \rightarrow N {a}_{3} X + 3 {H}_{2} O$ $\textcolor{red}{\left(1\right)}$

We can work out the number of moles of NaOH:

$n N a O H = 0.1 \times \frac{15}{1000} = 0.0015$

Because you need 3 moles of NaOH for every mole of ${H}_{3} X$ there will be some ${H}_{3} X$ left in excess.

Initial moles ${H}_{3} X = 0.001$

No. moles ${H}^{+}$ available = $0.001 \times 3 = 0.003$

So no. moles left after neutralisation = $0.003 - 0.0015 = 0.0015$

This means the no. moles ${H}_{3} X$ left must be $\frac{0.0015}{3} = 0.0005$

From $\textcolor{red}{\left(1\right)}$ the number of moles of ${X}^{-}$= (no. moles NaOH)/3=

$\frac{0.0015}{3} = 0.0005$

So we are left with a solution effectively consisting of ${H}_{3} X$ and ${X}^{3 -}$

${H}_{3} X r i g h t \le f t h a r p \infty n s 3 {H}^{+} + {X}^{-}$

${K}_{a \left(123\right)} = \frac{{\left[{H}^{+}\right]}^{3} \left[{X}^{-}\right]}{\left[{H}_{3} X\right]}$ $\textcolor{red}{\left(2\right)}$

${K}_{a \left(123\right)}$ is the product of the individual ${K}_{a}$ values:

${K}_{a \left(123\right)} = {K}_{a \left(1\right)} {K}_{a \left(2\right)} {K}_{a \left(3\right)}$

${K}_{a \left(123\right)} = \left(7.5 \times {10}^{- 4}\right) \times \left(1.7 \times {10}^{- 5}\right) \times \left(4 \times {10}^{- 7}\right)$

$= 5.1 \times {10}^{- 15}$

From $\textcolor{red}{\left(2\right)}$ $\Rightarrow$

${\left[{H}^{+}\right]}^{3} = \frac{{K}_{a \left(123\right)} \left[{H}_{3} X\right]}{\left[{X}^{-}\right]}$

${\left[{H}^{+}\right]}^{3} = \frac{{K}_{a \left(123\right)} \left[0.0005\right]}{\left[0.0005\right]}$

So:

${\left[{H}^{+}\right]}^{3} = {K}_{a \left(123\right)} \times 1 = 5.1 \times {10}^{- 15}$

From which:

$\left[{H}^{+}\right] = 1.72 \times {10}^{- 5}$

$p H = - \text{log} \left(1.72 \times {10}^{- 5}\right) = 4.76$