Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1678 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 titrating it with the standard solution. The endpoint was reached after the addition of 19.88 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

Answer 1

Answer:

Concentration of original solution = 1.66

Explanation:

We know that

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

We have given concentration of NaOH = 0.1678

Volume of NaOH = 19.88 mL = 0.01988 L

So moles of NaOH = volume x concentration of NaOH

= $0.01988\times 0.1678=0.00333mole$

Moles of $H_2SO_4$ in 10 mL of diluted solution = 1/2 x moles of NaOH

= $\frac{1}{2}$ x 0.00333 = 0.00166 mol

Moles of $H_2SO_4$ in 25 mL of original solution

= moles of H2SO4 in 250 mL of diluted solution

= $\frac{250}{10}$ x 0.00166 = 0.0415 mol

Concentration of original solution = $\frac{moles}{volume}$

= $\frac{0.0415}{0.025}=1.66$