Question

0.1 mole of H3POx is completely neutralised by 5.6 g KOH then x =?

Answer 1

X= 4
As in H3PO4 there are 3OH attatched to P(and one O double bonded with P) and the three OH donates 3H+ (it acts as an acid here) which is nutralised by 3 OH- supplied by KOH. Hence, it will be 4.

Answer 2

Answer: The value of x is 2

Explanation:

$H_3PO_x+3KOH\rightarrow K_3PO_x+3H2O$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $KOH=\frac{5.6g}{56g/mol}=0.1moles$

According to stochiometry:

1 mole of $H_3PO_x$ reacts with 3 moles of $KOH$

Now 0.1 mole of $H_3PO_4$ would require 0.3 mole of $KOH$ if all the $H^+$ are ionizable.

This means there is only one ionizable $H^+$ in $H_3PO_x$

Thus value of x is 2 as $H_3PO_2$ has basicity of 1.