Question

(6) 30.0 mL of 0.512 M sulfuric acid was added to 25.0 mL of 0.666 M sodium
hydroxide.
a. Write the balanced chemical equation.
b. Determine the molarity of either acid or base remaining after the reaction.
c. Find the molarity if a 50.00 mL aliquot of this solution is diluted to 250. mL.

Answer 1

Answer on Question #54742 – Chemistry – General Chemistry

Task:

30.0 mL of 0.512 M sulfuric acid was added to 25.0 mL of 0.666 M sodium hydroxide.

a. Write the balanced chemical equation.

b. Determine the molarity of either acid or base remaining after the reaction.

c. Find the molarity if a 50.00 mL aliquot of this solution is diluted to 250 mL.

Answer:

a. $\mathrm{H}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaOH} = \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{H}_{2} \mathrm{O}$

b. According to the equation, $n(\mathrm{NaOH}) = 2 \cdot n(\mathrm{H}_2\mathrm{SO}_4)$

$$n = C _ {M} \cdot V$$ $$n \left(H _ {2} S O _ {4}\right) = 0. 0 3 0 \cdot 0. 5 1 2 = 0. 0 1 5 \mathrm{~mol}$$ $$n (N a O H) = 0. 0 2 5 \cdot 0. 6 6 6 = 0. 0 1 7 \mathrm{~mol}$$

At the same time, the required amount of NaOH is: $n(NaOH)^{*} = 0.1536 \cdot 2 = 0.3072$ mol

So that in this situation NaOH is used fully, and the amount of $\mathrm{H}_2\mathrm{SO}_4$ left is:

$$n \left(H _ {2} S O _ {4}\right) _ {l e f t} = 0. 0 1 5 - \frac {0 . 0 1 7}{2} = 0. 0 0 7 \mathrm{~mol}$$

New molarity of that sulfuric acid will be:

$$C _ {M} \left(H _ {2} S O _ {4}\right) _ {l e f t} = \frac {0 . 0 0 7}{0 . 0 3 + 0 . 0 2 5} = 0. 1 2 7 \mathrm{~M}$$

c. A 50.00 mL aliquot of this solution contains sulfuric acid:

$$n \left(H _ {2} S O _ {4}\right) _ {\text {a l i q u o t}} = \frac {0 . 0 0 7}{5 5} \cdot 5 0 = 0. 0 0 6 4 \mathrm{~mol}$$

If it is diluted to $250\mathrm{ml}$, new molarity will be:

$$C _ {M} \left(H _ {2} S O _ {4}\right) _ {\text {a l i q u o t}} = \frac {0 . 0 0 6 4}{0 . 2 5} = 0. 0 2 5 \mathrm{~M}$$

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