# 20.00 ml of 0.10 M sodium carbonate is placed in a flask with phenolphthalein and methyl orange indicators present. 0.10 M hydrochloric acid is placed in the burette and run in slowly. How do you do the following calculations ?

## Calculate the pH at the following points: (a) 0.05 ml before the 1st end point. (b) 0.05 ml after the 1st end point. (c) 0.05 ml before the 2nd end point. (d) 0.05 ml past the second end point.

Mar 5, 2016

(a) pH = 9.02

(b) pH = 8.97

(c) pH = 5.06

(d) pH = 3.7

#### Explanation:

(A)

Carbonic acid is a weak diprotic acid with 2 ionisations:

${H}_{2} C {O}_{3} r i g h t \le f t h a r p \infty n s H C {O}_{3}^{-} + {H}^{+} \text{ "K_(a1)=4.27xx10^(-7)"mol/l}$

$H C {O}_{3}^{-} r i g h t \le f t h a r p \infty n s C {O}_{3}^{2 -} + {H}^{+} \text{ "K_(a2)=4.68xx10^(-11)"mol/l}$

In the flask we have $20.00 \text{ml}$ of $0.10 \text{M}$ sodium carbonate. As the acid is run in from the burette the following reaction occurs as the carbonate ion becomes protonated:

$C {O}_{3}^{2 -} + {H}^{+} \rightarrow H C {O}_{3}^{-}$

The initial number of moles of $C {O}_{3}^{2 -}$ is given by:

$n C {O}_{3}^{2 -} = 20.00 \times \frac{0.10}{1000} = 2.00 \times {10}^{- 3}$

At $0.05 \text{ml}$ before the end-point we have added $19.05 \text{ml}$ of $0.10 {\text{M"color(white)xHCl}}_{\left(a q\right)}$.

The number of moles of $H C l$ added is given by:

$n H C l = 19.05 \times \frac{0.10}{1000} = 1.905 \times {10}^{- 3}$

This will be equal to the number of moles of $H C {O}_{3}^{-}$ formed.

So the number of moles of $C {O}_{3}^{2 -}$ remaining is given by:

$n C {O}_{3}^{2 -} = \left(2.00 \times {10}^{- 3} - 1.905 \times {10}^{- 3}\right) = 0.095 \times {10}^{- 3}$

To get the number of moles of ${H}^{+}$ released we can use an ICE table based on $\text{mol/l}$

$\textcolor{w h i t e}{\times x} H C {O}_{3}^{-} \textcolor{w h i t e}{\times \times x} r i g h t \le f t h a r p \infty n s \textcolor{w h i t e}{\times \times} C {O}_{3}^{2 -} \textcolor{w h i t e}{\times \times x} + \textcolor{w h i t e}{\times \times x} {H}^{+}$

$\textcolor{red}{\text{I}} \textcolor{w h i t e}{\times x} 1.905 \times {10}^{- 3} \textcolor{w h i t e}{\times \times \times} 0.095 \times {10}^{- 3} \textcolor{w h i t e}{\times \times \times \times} 0$

$\textcolor{red}{\text{C}} \textcolor{w h i t e}{\times \times} - x \textcolor{w h i t e}{\times \times \times \times \times \times \times} + x \textcolor{w h i t e}{\times \times \times \times x} + x$

$\textcolor{red}{\text{E}} \textcolor{w h i t e}{\times x} \left(1.905 \times {10}^{- 3}\right) - x \textcolor{w h i t e}{\times x} \left(0.0905 \times {10}^{- 3}\right) + x \textcolor{w h i t e}{\times} x$

Since $x$ is much smaller than $n H C {O}_{3}^{-}$ and $n C {O}_{3}^{2 -}$ we can say that at equilibrium:

$\therefore n H C {O}_{3}^{-} = \left(1.905 \times {10}^{- 3}\right) - x \Rightarrow \left(1.905 \times {10}^{- 3}\right)$

and:

$n C {O}_{3}^{2 -} = \left(0.095 \times {10}^{- 3}\right) + x \Rightarrow \left(0.095 \times {10}^{- 3}\right)$

I will make this approximation in the rest of the calculations.

The expression for ${K}_{a 2}$ is:

${K}_{a 2} = \frac{\left[C {O}_{3}^{2 -}\right] \left[{H}^{+}\right]}{\left[H C {O}_{3}^{-}\right]}$

Since the total volume is common to all species we can write:

:.[H^+]=4.68xx10^(-11)xx(1.905xxcancel10^(-3))/(0.095xxcancel(10^(-3))

$\therefore \left[{H}^{+}\right] = 9.38 \times {10}^{- 10}$

$\therefore p H = - \log \left(9.38 \times {10}^{- 10}\right)$

color(red)(pH=9.02

This represents the pH just before the 1st end-point.

(B)

At the end-point the 1st protonation is complete:

$C {O}_{3}^{2 -} + {H}^{+} \rightarrow H C {O}_{3}^{-}$

So we now have $40.00 \text{ml}$ of a solution containing $2.00 \times {10}^{- 3}$ moles of $H C {O}_{3}^{-}$ ions.

As we continue to add acid the 2nd protonation starts to happen:

$H C {O}_{3}^{-} + {H}^{+} \rightarrow {H}_{2} C {O}_{3}$

After the end point at $20.00 \text{ml}$ we add $0.05 \text{ml}$ of $0.1 \text{M"color(white)(x)"HCl}$:

The number of moles of ${H}^{+}$ added is given by:

$n {H}^{+} = 0.05 \times \frac{0.10}{1000} = 5.00 \times {10}^{- 6}$

From the equation of the reaction for this stage of the titration we can say that:

$n {H}_{2} C {O}_{3}$ formed $= 5.00 \times {10}^{- 6}$

So the number of moles of $H C {O}_{3}^{-}$ remaining is given by:

$n H C {O}_{3}^{-} = \left(2.00 \times {10}^{- 3}\right) - \left(5.00 \times {10}^{- 6}\right) = 0.001995$

${K}_{a 1} = \frac{\left[H C {O}_{3}^{-}\right] \left[{H}^{+}\right]}{\left[{H}_{2} C {O}_{3}\right]}$

$\therefore \left[{H}^{+}\right] = {K}_{a 1} \times \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]}$

$\therefore \left[{H}^{+}\right] = 4.27 \times {10}^{- 7} \times \frac{5.00 \times {10}^{- 6}}{0.001995} = 1.0675 \times {10}^{- 9}$

$\therefore p H = - \log \left(1.0675 \times {10}^{- 9}\right)$

$\textcolor{red}{p H = 8.97}$

Phenolphthalein is an indicator that changes colour over the pH range 8.3 - 10 so is a suitable indicator for this end-point.

(C)

Now the titration continues to $0.05 \text{ml}$ short of the 2nd end-point which occurs at $40.00 \text{ml}$ of $\text{HCl}$ added.

Since the 1st end-point we have added $19.05 \text{ml}$ of $0.10 \text{M"color(white)(x)" HCl}$.

$\therefore n {H}^{+}$ added $= 19.05 \times \frac{0.10}{1000} = 1.905 \times {10}^{- 3}$

$\therefore n {H}_{2} C {O}_{3}$ formed $= 1.905 \times {10}^{- 3}$

$\therefore n H C {O}_{3}^{-}$ remaining = $\left(2.00 \times {10}^{- 3}\right) - \left(1.905 \times {10}^{- 3}\right) = 0.095 \times {10}^{- 3}$

${K}_{a 1} = \frac{\left[{H}^{+}\right] \left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}$

$\therefore \left[{H}^{+}\right] = {K}_{a 1} \times \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]}$

:.[H^+]=4.27xx10^(-7)xx(1.905xxcancel(10^(-3)))/(0.095xxcancel(10^(-3)

$\left[{H}^{+}\right] = 8.561 \times {10}^{- 6}$

$p H = - \log \left(8.561 \times {10}^{- 6}\right)$

$\textcolor{red}{p H = 5.06}$

(D)

When we get to the end point this reaction has reached completion:

$N {a}_{2} C {O}_{3 \left(a q\right)} + 2 H C {l}_{\left(a q\right)} \rightarrow {H}_{2} C {O}_{3 \left(a q\right)} + 2 N a C {l}_{\left(a q\right)}$

So we have $60 \text{ml}$ of a solution that contains $2 \times {10}^{- 3}$ moles of ${H}_{2} C {O}_{3}$.

We can find the pH of this solution. I will ignore the 2nd ionisation because ${K}_{a 2}$ is much smaller than ${K}_{a 1}$.

${K}_{a 1} = \frac{\left[H C {O}_{3}^{-}\right] \left[{H}^{+}\right]}{\left[{H}_{2} C {O}_{3}\right]}$

Since $\left[{H}^{+}\right] = \left[H C {O}_{3}^{-}\right]$ we can write:

${\left[{H}^{+}\right]}^{2} = {K}_{a 1} \times \left[{H}_{2} C {O}_{3}\right]$

We can find $\left[{H}_{2} C {O}_{3}\right]$ using $c = \frac{n}{v} :$

$\left[{H}_{2} C {O}_{3}\right] = \frac{2.00 \times {10}^{- 3}}{\frac{60.00}{1000}} = 0.033 \text{mol/l}$

$\therefore {\left[{H}^{+}\right]}^{2} = 4.27 \times {10}^{- 7} \times 0.033$

${\left[{H}^{+}\right]}^{2} = 1.41 \times {10}^{- 8}$

$\therefore \left[{H}^{+}\right] = 1.187 \times {10}^{- 4} \text{mol/l}$

$p H = - \log \left(1.187 \times {10}^{- 4}\right)$

$p H = 3.92$

Now we add an extra $0.05 \text{ml}$ of $0.1 \text{M"color(white)(x)" HCl}$. The extra number of moles of ${H}^{+}$ added is given by:

$0.10 \times \frac{0.05}{1000} = 5.00 \times {10}^{- 6}$

The number of moles of ${H}^{+}$ which are already there due to the dissociation of ${H}_{2} C {O}_{3}$ is given by:

$n = c \times v = 1.187 \times {10}^{- 4} \times \frac{60.00}{1000} = 7.122 \times {10}^{- 6}$

So we add this to the extra ${H}^{+}$ added to get the total moles of ${H}^{+}$ present:

$n {H}^{+} = \left(7.122 + 5\right) \times {10}^{- 6}$

$n {H}^{+} = 1.2122 \times {10}^{- 5}$

The new total volume is $60.05 \text{ml} .$

$\therefore \left[{H}^{+}\right] = \frac{1.2122 \times {10}^{- 5}}{\frac{60.05}{1000}} = 2.00 \times {10}^{- 4} \text{mol/l}$

$p H = - \log \left(2 \times {10}^{- 4}\right)$

$\textcolor{red}{p H = 3.7}$

So you can see the pH has dropped slightly.

I have ignored any ${H}^{+}$ from the ionisation of water.

The pH curve looks like this: (In our example A is at 20.00ml, B is at 40.00ml)

Methyl Orange is an indicator which changes colour over the range pH 3.1 - 4.4 so is suitable for this end-point.

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