A 10.0 ml of 0.121M H2SO4 is neutralized by 17.1 ml of KOH solution. what is the

Question

A 10.0 ml of 0.121M H2SO4 is neutralized by 17.1 ml of KOH solution. what is the molarity of the KOH solution is ____?

I solved this as

10.0ml of H2SO4 * 1L H2SO4/1000mLH2SO4*0.121M H2SO4/1L H2SO4*1mole KOH/1mole H2SO4, and divided it 0.0171L KOH. It's 0.0708 M KOH.

and I looked it up on online and some answers show 0.0708 but most of other answers show 0.142M.

I know its ratio is 1:2, but there's no balanced equation in the question. Do I have to multiply 2 although it doesn't show its balanced equation?

Explanation / Answer

Balanced chemical equation is:

H2SO4 + 2 KOH ---> K2SO4 + 2 H2O

Here:

M(H2SO4)=0.121 M

V(H2SO4)=10.0 mL

V(KOH)=17.1 mL

According to balanced reaction:

2*number of mol of H2SO4 =1*number of mol of KOH

2*M(H2SO4)*V(H2SO4) =1*M(KOH)*V(KOH)

2*0.121*10.0 = 1*M(KOH)*17.1

M(KOH) = 0.1415 M

Answer: 0.1415 M

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