Question

15.05 ml of a KOH solution neutralizes 25.6 mL of a 5.0 M H2SO4 solution. What is the concentration of the KOH solution?

Answer 1

The concentration of KOH solution is 17.009 moles/litre or 17.009M .

Step-by-step explanation:

Here , the given reaction is -

`2KOH +H_(2)SO_(4)$\rightarrow$K_(2)SO_(4)+2H_(2)O`

2moles of KOH requires 1mole for complete neutralisation.

Therefore ,

2 x number of moles of KOH = 1 x number of moles of
`H_(2)SO_(4)`

2 x (molarity of KOH x volume of KOH) = (Molarity of
`H_(2)SO_(4)` x volume of

`H_(2)SO_(4)`)

1 mole of
`OH^- $\rightarrow$` 1mole of
`H^+`

Molarity (KOH) = concentration (c)

volume (KOH) =15.05ML

Molarity (
`H_(2)SO_(4)` ) = 5.0 M

Volume of (
`H_(2)SO_(4)` ) = 25.6ML

`(1*(C *15.05ml))/(OH^-) = ((2* 25.6 * 5))/(H^+)`

C =
`(0.2*25.6*5)/(15.05)`

C= 17.009 moles/litre

Hence , the concentration of KOH solution is 17.009 M or moles/litre .