Question

$H_3{PO}_4$ is a tribasic acid and one of its salt is $Na{H}_2{PO}_4$. What volume of 1M NaOH solution should be added to 12g $Na{H}_2{PO}_4$ to convert it into $N{a}_3{PO}_4$? (at.wt of P=31)

• A. 100 mL
• B. 200 mL
• C. 80 mL
• D. 300 mL

Answer 1

The correct option is B 200 mL

$Na{H}_{2}P{O}_4$ + $2NaOH$ $\to N{a}_{3}P{O}_4+H_{2}O$

From the above reaction we can say that,

1 mol of $Na{H}_{2}P{O}_4$ reacts with 2 mol of NaOH.

Molar mass of $Na{H}_{2}P{O}_4$ = 120 g / mol

12g of $Na{H}_{2}P{O}_4$ contains = 0.1 mol

So the number of moles of NaOH required = 0.2 mol

Molarity of NaOH = 1M

1 mol = 1000 ml

0.2 mol = ?

So, the volume of 1M NaOH solution = 200 ml