How much volume of 0.1 M H3PO3 (dibasic acid) solution is required for complete reaction with
400 ml, 0.3 M KOH solution?
Answers
Answer: 1 mol of the acid will be held in: 10 L of the solution
Explanation:
For getting the answer by applying the direct formula, put:
N1V1 = N2V2, where N is normality, V is volume of the substance.
=> n1 M1 V1 = n2 M2 V2 , where n is the valence factor (here, number of H+ ions liberated by acid or OH- ions by the base), and M is the molarity.
Assuming 1 to be NaOH and 2 to be H2SO4, we have:
1 X 0.5 X 1 = 2 X 0.1 X V2
=> V2 = 2.5 L.
Easy-peasy, lemon-squeezy!
In case you forgot the formula, but understand the fundamentals thoroughly, go on with this method:
Amount of NaOH in 1 L 0.5 M NaOH = 0.5 mol, as molarity is nothing but the number of moles present per litre.
Now, the reaction between NaOH and H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
Thus, 1 mol NaOH needs 0.5 mol H2SO4.
Here, we need 0.5 X 0.5 = 0.25 mol of the acid.
Now, 1 L of the acid solution has : 0.1 mol of the acid.
=> 1 mol of the acid will be held in: 10 L of the solution
=> 0.25 mol of the acid (the amount we require) will be held in: 10 X 0.25 L = 2.5 L of the solution.
Answer:
Apply the Law of equivalence we get,
equivalence of acid = equivalence of base
0.1×V=0.04×1
V=0.4 L=0.4×1000=400 ml
Hence, the correct option is A