Question

A student reacts 25.0 mL of 0.175 M H3PO4 with 25.0 mL of 0.205 M KOH. Write a balanced chemical equation to show this reaction. Calculate the concentrations of H3PO4 and KOH that remain in solution, as well as the concentration of the salt that is formed during the reaction.

Answer 1

Answer: The concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(1)

For KOH:

Initial molarity of KOH solution = 0.205 M

Volume of solution = 25.0 mL = 0.025 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

`0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L* 0.025L)=5.123* 10^(-3)mol`

For phosphoric acid:

Initial molarity of phosphoric acid solution = 0.175 M

Volume of solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

`0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L* 0.025L)=4.375* 10^(-3)mol`

The chemical equation for the reaction of KOH and phosphoric acid follows:

`3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O`

By Stoichiometry of the reaction:

3 moles of KOH reacts with 1 mole of phosphoric acid

So,
`5.123* 10^(-3)` moles of KOH will react with =
`(1)/(3)* 5.123* 10^(-3)=1.708* 10^(-3)mol` of phosphoric acid

As, given amount of phosphoric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, KOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of phosphoric acid =
`(4.375-1.708)* 10^(-3)=2.667* 10^(-3)mol`

By Stoichiometry of the reaction:

3 moles of KOH produces 1 mole of potassium phosphate

So,
`5.123* 10^(-3)` moles of KOH will produce =
`(1)/(3)* 5.123* 10^(-3)=1.708* 10^(-3)moles` of potassium phosphate

• For potassium phosphate:

Moles of potassium phosphate =
`1.708* 10^(-3)moles`

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

`\text{Molarity of potassium phosphate}=(1.708* 10^(-3))/(0.050)=0.0342M`

• For phosphoric acid:

Moles of excess phosphoric acid =
`2.667* 10^(-3)moles`

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

`\text{Molarity of phosphoric acid}=(2.667* 10^(-3))/(0.050)=0.0533M`

• For KOH:

Moles of KOH remained = 0 moles

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

`\text{Molarity of KOH}=(0)/(0.050)=0M`

Hence, the concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.