Question

0.01 mole of $H_3{PO}_x$ is completely neutralised by 0.56 gram of KOH hence :

• A. x = 3 and given acid is diabasic
• B. x = 2 and given acid is mono basic
• C. x = 3 and given acid is mono basic
• D. x = 4 and given acid forms three series of salt

Answer 1

Answer: (B)

x = 2 and given acid is mono basic

$H_{3}P{O}_x+3KOH⟶$ $K_{3}P{O}_x+3H_{2}O$

Moles of $KOH$ = $\frac{0.56}{56}=0.01moles$

According to stochiometry:

1 mole of $H_{3}P{O}_x$ react with 3 moles of$KOH$

Now 0.01 mole of $H_{3}P{O}_x$ will require 0.03 mole of $KOH$ if all $H^{+}$ are ionizable

This means there is only one ionizable $H^{+}$ in $H_{3}P{O}_x$

Thus value of x is 2 as $H_{3}P{O}_2$ has basicity 1.

Hence,option B is correct answer.