Question

In a titration of H2 SO4 with NaOH, 60mL of 0.02 M NaOH was needed to neutralize 15mL of H2 SO4 . What is the molarity of the acid?

Answer 1

Solution.

$V(NaOH)=60mL;$

$M(NaOH)=0.02M;$

$V(H_2SO_4) =15mL;$

$M(H_2SO_4)-?$ ;

Because the reaction is a 1:1 rxn ratio, it is useful to apply

(Molarity x Volume)Acid = (Molarity x Volume)Base;

$M(H_2SO_4)\sdot V(H_2SO_4)=M(NaOH)\sdot V(NaOH) \implies$

$M(H_2SO_4)=\dfrac{M(NaOH)\sdot V(NaOH)}{V(H_2SO_4)};$

$M(H_2SO_4)=\dfrac{0.02M\sdot 60mL}{15mL}= 0.08M;$

Answer: $M(H_2SO_4)=0.08M.$