Question

A 25.00 mL sample of an H2SO4 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 H2SO4 solution? Express your answer in molarity to four significant figures.

Answer 1

`0.1090\ \text{M}`

Step-by-step explanation:

Volume of
`KOH` = 41.22 mL

Molarity of
`KOH` = 0.1322 M

Number of moles of
`KOH`

`n=\text{Molarity}* \text{Volume of }KOH\\\Rightarrow n=0.1322* 41.22* 10^(-3)\\\Rightarrow n=0.005449284\ \text{mol}`

2 moles of
`KOH` reacts with 1 mole of
`H_2SO_4`

So,

`0.005449284\ \text{mol}` of
`KOH` reacts with M mole of
`H_2SO_4`

`M=(0.005449284)/(2)=0.002724642\ \text{mol}`

Volume of
`H_2SO_4=25* 10^(-3)\ \text{L}`

Concentration of
`H_2SO_4` is

`M=(0.002724642)/(25* 10^(-3))\\\Rightarrow M=0.10898568\approx 0.1090\ \text{M}`

The concentration of the unknown
`H_2SO_4` solution is
`0.1090\ \text{M}`.