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Solved: A 25.00-mL sample of an H2SO4 solution of unknown concentration is titrated with

Chapter 14, Problem 49P

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QUESTION:

A 25.00-mL sample of an \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of unknown concentration is titrated with a 0.1322 M KOH solution. A volume of 41.22 mL of KOH is required to reach the equivalence point. What is the concentration of the unknown \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

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QUESTION:

A 25.00-mL sample of an \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of unknown concentration is titrated with a 0.1322 M KOH solution. A volume of 41.22 mL of KOH is required to reach the equivalence point. What is the concentration of the unknown \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

ANSWER:

Step 1 of 3

The reaction which takes place between KOH and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is as follows.

\(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

From the balanced equation , 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is required to  neutralize 2 moles of KOH

So, 1 mol \(\mathrm{H}_{2} \mathrm{SO}_{4}\) = 2 mol KOH

The conversion factor will be \(\frac{1\mathrm{\ mol}\mathrm{\ H}_2\mathrm{SO}_4}{2\mathrm{\ mol}\mathrm{\ KOH}}\)

The conversion factor of “L” into “mL” : \(\frac{1\ L}{1000\ mL}\)

From the given,

\(\begin{array}{l} \text { Molarity of } \mathrm{KOH}=0.1322 \mathrm{mol} \\ \text { Molarity of } \mathrm{KOH}=0.1322 \mathrm{M}=0.1322 \mathrm{mol} / \mathrm{L} \\ \text { The conversion factor will be: } \quad \frac{0.1322 \mathrm{molKOH}}{1 \mathrm{LKOH}} \end{array}\)

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