Question

A titration is performed to determine the amount of sulfuric acid, H2SO4, in a 6.5 mL sample taken from car battery. About 50 mL of water is added to the sample, and then it is titrated with 43.37 mL of a standard 0.5824 molar NaOH solution. You balanced this reaction in a previous problem. How what is the molar concentration of sulfuric acid in the original sample

Answer 1

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Explanation:

To calculate the molarity 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:

$n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL$

Putting values in above equation, we get:

$2\times M_1\times 56.5=1\times 0.5824\times 43.37$

$M_1=0.2235$

Now to calculate the molarity of original solution:

$M_1\times 6.5=0.2235\times 56.5$

$M_1=1.943$

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M