1.How many milliliters of 0.0050 N KOH are required to neutralize 53 mL of 0.005

Question

1.How many milliliters of 0.0050 N KOH are required to neutralize 53 mL of 0.0050 N H2SO4 ?

2.How many milliliters of 0.0050 N KOH are required to neutralize 53 mL of 0.0050 M H2SO4?

3.What is the molarity of a solution made by dissolving 3.5 g of Ca(OH)2 in enough water to make 200.0 ml of solution?What is its normality?

4.How many milliliters of 0.50M NaOH solution are required to titrate 20.0 mL of a 0.30 M H2SO4 solution to an end point?

5. How many milliliters of 0.250 M NaOH are required to neutralize 50.0 mL of 0.400 M H2SO4 ? The balanced neutralization reaction is: H2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l).

Explanation / Answer

1)
for neutralisation,
equivalent of KOH = Equivalent of H2SO3
N(KOH)*V(KOH) = N(H2SO4)*V(H2SO4)
0.0050*V(KOH) = 0.0050*53
V(KOH) = 53 mL
Answer: 53 mL
2)
for neutralisation,
equivalent of KOH = Equivalent of H2SO3
N(KOH)*V(KOH) = N(H2SO4)*V(H2SO4)
1*M(KOH)*V(KOH) = 2*M(H2SO4)*V(H2SO4)
0.0050*V(KOH) = 2*0.0050*53
V(KOH) = 106 mL
Answer: 106 mL

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