The balanced equation for the neutralization reaction of aqueous H2SO4 with aqueous KOH is shown.
H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2 SO4(aq)
What volume of 0.500M KOH is needed to react completely with 17.2mL of 0.100M H2SO4?
Balanced chemical equation is:
H2SO4 + 2 KOH ---> K2SO4 + 2 H2O
Here:
M(H2SO4)=0.1 M
M(KOH)=0.5 M
V(H2SO4)=17.2 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*0.1 M *17.2 mL = 1*0.5M *V(KOH)
V(KOH) = 6.88 mL
Answer: 6.88 mL
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