Question

We find that 18.90 milliliters of a 2.50M KOH solution are required to titrate 35.0 milliters of a H2S04 solution. What is the molarity of the H2S04 solution? 2 2KOH + H2S04 K2S04 2 H20 A. 0.675 M B. 1.57 M C. 0.0711 M D. 0.128 M E. 4.04 M F. 0.903 M G. 1.18 M H. 0.0173 M

Answer 1

Concentration of
`H_(2)SO_(4)` is 0.675 M

Step-by-step explanation:

According to balanced equation, 2 moles of KOH neutralizes 1 mol of
`H_(2)SO_(4)`.

18.90 mL of 2.50M KOH =
`(2.50* 18.90)/(1000)moles` of KOH = 0.04725 moles of KOH

If molarity of
`H_(2)SO_(4)` is C (M) then-

moles of
`H_(2)SO_(4)` are neutralized =
`(C* 35.0)/(1000)moles` of
`H_(2)SO_(4)`

Hence,
`(1)/(2)* 0.04725=(C* 35.0)/(1000)`

or, C = 0.675

So, concentration of
`H_(2)SO_(4)` is 0.675 M