A 1.50 mL sample of a sulphuric acid solution from an automobile storage battery is titrated with 1.47 M odium hydroxide solution to a phenolphthalein endpomt, requiring 23.70 ml. What is the molarity of the sulphuric acid solution?

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23.2 M

11.6 M

0.181 M

6.30 M

answer is B.

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Detailed Solution

There is a neutralization reaction happening.

$H_{2} S O_{4} (a q) + 2 N a O H (a q) \to N a_{2} S O_{4} (a q) + 2 H_{2} O (l)$

We use the equation provided by the neutralisation reaction to get the acid concentration:

$n_{1} M_{1} V_{1} = n_{2} M_{2} V_{2}$

where,
$n_{1}, M_{1}$ and $V_{1}$ are the $H_{2} {SO}_{4}$ acid's n-factor, molarity, and volume.
$n_{2}, M_{2} and V_{2}$ are the base's $NaOH$ molarity, n-factor, and volume.

We are given:

$\begin{aligned} n_{1} = 2 \\ M_{1} = ? \\ V_{1} = 1.50 mL \\ n_{2} = 1 \\ M_{2} = 1.47 M \\ V_{2} = 23.70 mL \end{aligned}$

Putting values in above equation, we get:

$\begin{aligned} 2 \times M_{1} \times 1.50 mL = 1 \times 1.47 M \times 23.70 ml \\ M_{1} = \frac{1 \times 1.47 M \times 23.70 ml}{2 \times 1.50 ml} = 11.613 M \end{aligned}$