Question

What is the molarity of a KOH solution if 28.2 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of the KOH solution?

Answer 1

1) Chemical equation

`2KOH_{}+H_2SO_4\rightarrow K_2SO_4+2H_2O_{}_{}`

2) Moles of H2SO4 that reacted

`M=\frac{moles\text{ of solute}}{\text{liters of solution}}`
`n=M\cdot v`

Convert the units of volume

`L=28.2mL\cdot(1L)/(1000mL)0.0282L`
`mol_{}H_2SO_4=0.220M\cdot0.0282L=0.006204molH_2SO_4_{}`

3) Moles of KOH that reacted.

The molar ratio is 2 mol KOH: 1 mol H2SO4.

`mol_{}KOH=0.006204molH_2SO_4\cdot\frac{2molKOH_{}}{1molH_2SO_4}=0.012408molKOH_{}`

4) Molarity of KOH

Convert the units of volume

`L=25.0mL\cdot(1L)/(1000mL)=0.0250L`
`M=(n)/(v)`
`M=\frac{0.012408molKOH_{}}{0.0250L}=0.49632MKOH_{}`

The molarity of the KOH solution is 0.49632 M.

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