During a volumetric analysis experiment, 20.0\text{ cm}^3 of a 0.050\text{ mol dm}^{-3} so… — JAMB UTME

During a volumetric analysis experiment, $20.0\text{ cm}^3$ of a $0.050\text{ mol dm}^{-3}$ sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3$ ) solution was completely neutralized by $25.0\text{ cm}^3$ of a tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution. What is the mass concentration of the tetraoxosulfate(VI) acid solution in $\text{g dm}^{-3}$ ? $[\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]$

$3.92\text{ g dm}^{-3}$ Answer
B
$7.84\text{ g dm}^{-3}$
C
$1.96\text{ g dm}^{-3}$
D
$6.13\text{ g dm}^{-3}$

Answer

The mass concentration of the tetraoxosulfate(VI) acid solution is

3.92\text{ g dm}^{-3}

The balanced chemical reaction shows a

1:1

stoichiometric ratio between

\text{Na}_2\text{CO}_3

and

\text{H}_2\text{SO}_4

. Substituting the given values into the titration equation yields a molar concentration of

0.040\text{ mol dm}^{-3}

. Multiplying this molarity by the molar mass of

\text{H}_2\text{SO}_4

(

98.0\text{ g mol}^{-3}

) correctly yields

3.92\text{ g dm}^{-3}

Step-by-Step Solution

Write the balanced chemical equation for the neutralization reaction.

\text{Na}_2\text{CO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

Determines the mole ratio between the acid (

n_a = 1

) and base (

n_b = 1

Apply the volumetric neutralization formula to find the molar concentration of the acid (

C_a

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{C_a \times 25.0}{0.050 \times 20.0} = \frac{1}{1} \implies C_a = 0.040\text{ mol dm}^{-3}

Calculates the molarity of the tetraoxosulfate(VI) acid solution.

Calculate the molar mass of tetraoxosulfate(VI) acid (

\text{H}_2\text{SO}_4

\text{Molar Mass} = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-3}

Required to convert molar concentration to mass concentration.

Convert molar concentration to mass concentration.

\text{Mass concentration} = C_a \times \text{Molar Mass} = 0.040 \times 98.0 = 3.92\text{ g dm}^{-3}

Obtains the final concentration in grams per cubic decimetre.

Key Concept

Volumetric Analysis and Concentration Conversions

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