Question

$P_4{O}_6$ reacts with water according to equation $P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_{3-}$.
Calculate the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of $P_4{O}_{6}in{H}_{2}O$.

Answer 1

Calculating the moles of NaOH

Given reaction is:
$P_4{O}_6+6H_{2}O\to 4H_{3}P{O}_3$ --(i)

Thus, when $P_4{O}_6$ reacts with water, formation of phosphorous acid $H_{3}P{O}_3$ takes place.

Reaction of $H_{3}P{O}_3$ with NaOH (sodium hydroxide)
$4H_{3}P{O}_3+8NaOH\to 4N{a}_{2}HP{O}_3+8H_{2}O$ --(ii)

Now, 1.1 g of $P_4{O}_6$ is dissolved in $H_{2}O$

$\therefore$ Number of moles $=\frac{givenmass}{molarmass}=\frac{1.1}{220}=0.005mol$

According to reaction (i)
1 mole of $P_4{O}_6$ reacts to give $\to$ 4 mol of $H_{3}P{O}_3$
$\therefore$ 0.005 mol of $P_4{O}_6$ will give $\to 0.005\times 4\to$ 0.002 mol of $H_{3}P{O}_3$

Similarly, according to reaction (ii)
4 mol of $H_{3}P{O}_3$ reacts with $\to$ 8 mol of NaOH
Thus, 1 mol of $H_{3}P{O}_3$ will react with $=\frac{8}{4}=2molofNaOH$

Therefore, 0.02 mol of $H_{3}P{O}_3$ will react with = $0.02\times 2=0.04molofNaOH$
Thus, 0.04 mol of NaOH are required to neutralize $H_{3}P{O}_3$ acid.

Calculation Volume of 0.1 M NaOH

$Molarities=\frac{no.ofmoles}{Volumeofsolution\left(inL\right)}$

$Volume=\frac{no.ofmoles}{molarity}\Rightarrow \frac{0.04}{0.1}$

$=0.4L=400mL$

Thus, 400 mL of 0.1 M NaOH solution is required to neutralize the acid formed by dissolving 1.1 g of $P_4{O}_{6}in{H}_{2}O$.