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

11.613 M is the molarity of the sulfuric acid solution.

Step-by-step 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* M_1* 1.50 mL=1* 1.47 M* 23.70 ml`

`M_1=(1* 1.47 M* 23.70 ml)/(2* 1.50 ml)=11.613 M`

11.613 M is the molarity of the sulfuric acid solution.