Respuesta :
ANSWER
The molarity of the unknown phosphoric acid is 0.0762 moles
STEP-BY-STEP 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.
[tex]H_3PO_{4(aq)}\text{ + }3NaOH_{(aq)}\text{ }\rightarrow3H_2O_{(l)}\text{ + }Na_3PO_{4(aq)}[/tex]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
[tex]\frac{C_AV_A}{C_BV_B}\text{ = }\frac{n_A}{n_B}[/tex]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
[tex]\begin{gathered} \frac{C_{A\cdot\text{ }}20}{0.1028\cdot\text{ 44.7}}\text{ = }\frac{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{ = }\frac{4.573184}{60} \\ C_A\text{ = 0.0762 mole} \end{gathered}[/tex]
