Question

A student found that it required 44.7 mL of a 0.1028 M NaOH Solution to neutralize 20.0 mL of an unknown H3PO4 Solution which react according to the equation: 3 NaOH + H3PO4 = Na3PO4 + 3H2O. The molarity of the unknown phosphoric acid solution is:

Answer 1

The molarity of the unknown phosphoric acid is 0.0762 moles

Explanation

What to find? The molarity of the unknown phosphoric acid

Given parameters

Volume of base = 44.7mL

Concentration of base = 0.1028 mole

Volume of acid = 20.0mL

nA = 1

nB = 3

To find the concentration of the unknown phosphoric acid, we need to write the balanced equation for the reaction.

`H_3PO_(4(aq))\text{ + }3NaOH_((aq))\text{ }\rightarrow3H_2O_((l))\text{ + }Na_3PO_(4(aq))`

From the equation of the reaction; This means that 1 mole of H3PO4 neutralizes 3 moles of NaOH

To find the molarity of H3PO4, we will need to apply the below formula

`(C_AV_A)/(C_BV_B)\text{ = }(n_A)/(n_B)`

Where;

CA = Concentration of the acid

VA = Volume of acid

CB = Concentration of the base

VB = volume of the base

nA = mole ratio of acid

nB = mole ratio of base

Substitute the parameters into the above formula

`\begin{gathered} \frac{C_{A\cdot\text{ }}20}{0.1028\cdot\text{ 44.7}}\text{ = }(1)/(3) \\ Cross\text{ multiply} \\ 3\cdot C_A\cdot\text{ 20 = 0.10208 }\cdot\text{ 44.7 }\cdot\text{ 1} \\ 60C_A\text{ = 4.573184} \\ Divide\text{ both sides by }60 \\ \frac{60C_A}{60\text{ }}\text{ = }(4.573184)/(60) \\ C_A\text{ = 0.0762 mole} \end{gathered}`