Question

How many millilitres of 0.2M H₃PO₄ must be added to 100mL of 0.21M KOH to give a solution that has a concentration of 0.02M in H₃PO₄?

(1) 10mL

(2) 50mL

(3) 100mL

(4) 75mL

Answer 1

Correct option is (2) 50 mL

n-factor of H₃PO₄ = 3

n-factor of KOH = 1

Let say, Volume of H₃PO₄ = x mL

\(\therefore \) total volume of solution = (100 + x) mL

\(\therefore \) Number of milli equivalent of KOH = 1 \(\times\) 100 \(\times\) 0.21

= 21 milli equivalent of KOH.

\(\therefore \) Number of milli equivalent of H₃PO₄ = 3 \(\times\) x \(\times\) 0.2

= 0.6 x milli equivalent

After addition of KOH,

\(\therefore \) Number of milli equivalent of H₃PO₄ left = (0.6x - 21) milli equivalent.

\(\therefore \) Molarity of final solution = \(\frac{\text{Number of milli equivalent of }H_3PO_4}{\text{n-factor of }H_3PO_4 \times \text{Volume of solution in (mL)}}\)

\(0.02= \frac{(0.6 x- 21)}{3\times(x + 100)}\)

⇒ \(0.6x - 21 = 0.06x + 6\)

⇒ \(0.54 x = 27\)

\(x = 50 \,mL\) 

Hence, volume of H₃PO₄ will be 50 mL.