# A 0.3240g sample of impure Na_2CO_3 was dissolved in 50.00mL of 0.1280M HCl. The excess acid then requires 30.10mL of 0.1220M NaOH for complete neutralization. How do you calculate the % Na_2CO_3 (MM = 105.99) in the sample?

Jul 2, 2016

You can do it like this:

#### Explanation:

Sodium carbonate reacts with hydrochloric acid:

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

• We can find the number of moles of $\text{HCl}$ before the reaction.

• We then use the titration result to find the number of moles remaining after the reaction.

• By subtracting the two we can get the number of moles of $\text{HCl}$ which have reacted.

• From the equation we can find the number of moles of ${\text{Na"_2"CO}}_{3}$.

• From this we get the mass of ${\text{Na"_2"CO}}_{3}$.

• We can then work out the percentage purity.

Concentration = moles of solute / volume of solution.

$c = \frac{n}{v}$

$\therefore n = c \times v$

$\therefore$ Initial moles of sf(HCl"=0.1280xx50.00/1000=6.400xx10^(-3))

The acid remaining is titrated with $\text{NaOH}$:

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

$\textsf{n O {H}^{-} = 0.1220 \times \frac{30.10}{1000} = 3.6722 \times {10}^{- 3}}$

Since they react in a 1:1 ratio the no. of moles of $\text{HCl}$ must be the same:

$\textsf{n H C l = 3.6722 \times {10}^{- 3}}$

$\therefore$ The no. moles used up:

$\textsf{= \left(6.400 - 3.6722\right) \times {10}^{- 3} = 2.7228 \times {10}^{- 3}}$

From the original equation you can see that the no. moles of ${\text{Na"_2"CO}}_{3}$ must be half of this.

$\therefore \textsf{n N {a}_{2} C {O}_{3} = \frac{2.7228 \times {10}^{- 3}}{2} = 1.3614 \times {10}^{- 3}}$

$\textsf{{M}_{r} \left[N {a}_{2} C {O}_{3}\right] = 105.99}$

$\therefore$ $\textsf{m a s s \text{ "Na_2CO_3=1.3614xx10^(-3)xx105.99=0.14429" } g}$

$\therefore$ sf("percentage purity"" "=(0.14429)/(0.3240)xx100=44.53%)

This technique is known as a "back titration".

Jul 4, 2016

~~44.61%

#### Explanation:

The equivalent mass of $N {a}_{2} C {O}_{3}$
=("molar mass of "Na_2CO_3)/"total valency of metal"

$= \frac{105.99}{2} = 52.995 \frac{g}{\text{equivalent}}$

Let the amount of $N {a}_{2} C {O}_{3}$ in 0.3240g sample be $\left(x g\right) = \frac{x}{52.995} \text{g.equivalent}$

In HCl and NaOH the molar mass and eqivalent mass are same.So for their solutions molarity is same as normality.

So 50 mL 0.1280M or 0.1280N HCl solution will contain $\frac{50 \times 0.1280}{1000} \text{g.equivalent}$ of HCl

And 30.10mL 0.1220M or N NaOH solution will contain $\frac{30.1 \times 0.1220}{1000} \text{g.eqivalent}$ NaOH
Now by the law of equivalent proportion nutralisation will occur if total no.of g.eqivalent of base is same as total no.of g.equivalent of acid.

Hence

$\frac{x}{52.995} + \frac{30.1 \times 0.122}{1000} = \frac{50 \times 0.1280}{1000}$

$x = \frac{50 \times 0.128 - 30.1 \times 0.122}{1000} \times 52.995$
$\approx 0.14456 g$

$\text{% of purity } = \left(\frac{0.14456}{0.324}\right) \times 100 \approx 44.61$