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TEA [102]
3 years ago
6

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solutio

n. 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?
Chemistry
1 answer:
Alja [10]3 years ago
8 0

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

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