Question

A 1.50 mL sample of a sulfuric acid solution from an automobile storage battery is titrated with 1.47 M sodium hydroxide solution to a phenolphthalein endpoint, requiring 23.70 mL. What is the molarity of the sulfuric acid solution

Answer 1

Answer:

11.613 M is the molarity of the sulfuric acid solution.

Explanation:

The reaction taking place is neutralization reaction:$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=1.50 mL\\n_2=1\\M_2=1.47 M\\V_2=23.70 mL$

Putting values in above equation, we get:

$2\times M_1\times 1.50 mL=1\times 1.47 M\times 23.70 ml$

$M_1=\frac{1\times 1.47 M\times 23.70 ml}{2\times 1.50 ml}=11.613 M$

11.613 M is the molarity of the sulfuric acid solution.