Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1790 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 18.07 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

Answer 1

Answer: The concentration of original sulfuric acid solution is 1.62 M

Explanation:

Let the original concentration of sulfuric acid be 'x' M

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=xM\\V_1=25.00mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$x\times 25.00=M_2\times 250.0\\\\M_2=\frac{x\times 25.0}{250}=\frac{x}{10}$

Now, 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:

$n_1=2\\M_1=\frac{x}{10}M\\V_1=10.00mL\\n_2=1\\M_2=0.1790M\\V_2=18.07mL$

Putting values in above equation, we get:

$2\times \frac{x}{10}\times 10.00=1\times 0.1790\times 18.07\\\\x=\frac{1\times 0.1790\times 18.07\times 10}{2\times 10.00}=1.62M$

Hence, the concentration of original sulfuric acid solution is 1.62 M