Question

What volume of \(0.200MNaOH\)is necessary to neutralize the solution produced by dissolving \(2.00\;{g^{g }}PC{l_3}\) is an excess of water? Note that when \({H_3}P{O_3}\)is titrated under these conditions, only one proton of the acid molecule reacts.

Short answer

The \(73.0ml\) neutralisation, a certain amount of NaOH is necessary.

Step-by-step solution

Definition of NaOH

Sodium hydroxide (NaOH), often known as caustic soda or lye, is a versatile chemical that can be employed in a wide range of manufacturing processes. Chlorine produces sodium hydroxide as a by product.

Determine thevolume of \(0.200MNaOH\) is necessary to neutralize the solution produced by dissolving \(2.00\;{g^{g }}PC{l_3}\) is an excess of water 

Given Information:

Mass of \(PC{l_3} = 2.0\;g\)

Concentration of \(NaOH = 0.200M\)

The response of \(PC{l_3}\)with \({H_2}O\) can be expressed as follows:

\(PC{l_3} + 3{H_2}O \to {H_3}P{O_3} + 3HCl\)

Based on the stoichiometry of the reaction:

1 mole of \(PC{l_3}\) mixes with three moles \({H_2}O\) to make one mole of \({H_3}P{O_3}\)

The phosphoric acid that resulted \({H_3}P{O_3}\) when the base is the neutralisation reaction of titrated NaOH is as follows:

\({H_3}P{O_3} + NaOH \to Na{H_2}P{O_3} + {H_2}O\)

Thus, 1 mole of \({H_3}P{O_4}\) 1 mole of NaOH neutralises it.

Moles of \(PC{l_3}\)reacted \( = \frac{{2.0\;g}}{{137.33\;g - mo{l^{ - 1}}}} = 0.0146\) moles

Based on the stoichiometry of the reaction:

Moles of \({H_3}P{O_3}\) during titration =0.0146 moles = moles of NaOH neutralised

Volume of NaOHrequired \( = \frac{{ moles of NaOH}}{{ Molarity of NaOH}}\)

\(\begin{array}{} &= \frac{{0.0146 moles }}{{0.200 moles. {L^{ - 1}}}}\\ &= 0.073\;L\\ &= 73.0\;mL\end{array}\)

As a result, the amount of NaOH needed to neutralise \({H_3}P{O_3}\)formed is all over \(73.0ml\).