Question

A sample of oxalic acid dihydrate (126.07g/mL) with mass of 0.1473g was titrated by the addition of 35.87mL potassium hydroxide That solution of potassium hydroxide required 22.48mL to neutralize 10.00mL of nitric acid, determine molarity of the nitric acid.

Answer 1

Moles of oxalic acid = (0.1473/126.07) = 0.00117

Balanced reaction between oxalic acid and Potassium hydroxide.

H₂C₂O₄ .2 H₂O + 2 KOH $\rightarrow$ K₂C₂O₄ + 4 H₂O (l)

Hence, at neutralisation point ,

1 mole of oxalic acid completely reacts with 2 moles KOH.

Now, moles of oxalic acid taken = 0.00117

Then , moles of KOH needed = 2*0.00117 = 0.00234

Now, volume of KOH = 35.87 mL = 0.03587 L

Volume * molarity = mole

Or, 0.03587 * molarity = 0.00234

Or, molarity of KOH = (0.00234/0.03587) = 0.0652 M

Reaction between KOH and HNO₃

KOH + HNO₃ = KNO₃ + H₂O

At neutralization point,

Moles of KOH = moles of HNO₃

Therefore, M_acid * V_acid = M_base * V_base

V_acid = 10.00 mL

M_base = 0.0652 M

V_base = 22.48 mL

Then, M_acid * 10.00 = 0.0652*22.48

Or, M_acid = (1.46/10.00) = 0.146 M

Hence, molarity of nitric acid = 0.146