Respuesta :
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.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(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:
[tex]0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L\times 0.025L)=5.123\times 10^{-3}mol[/tex]
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:
[tex]0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L\times 0.025L)=4.375\times 10^{-3}mol[/tex]
The chemical equation for the reaction of KOH and phosphoric acid follows:
[tex]3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O[/tex]
By Stoichiometry of the reaction:
3 moles of KOH reacts with 1 mole of phosphoric acid
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}mol[/tex] 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 = [tex](4.375-1.708)\times 10^{-3}=2.667\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
3 moles of KOH produces 1 mole of potassium phosphate
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}moles[/tex] of potassium phosphate
- For potassium phosphate:
Moles of potassium phosphate = [tex]1.708\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of potassium phosphate}=\frac{1.708\times 10^{-3}}{0.050}=0.0342M[/tex]
- For phosphoric acid:
Moles of excess phosphoric acid = [tex]2.667\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of phosphoric acid}=\frac{2.667\times 10^{-3}}{0.050}=0.0533M[/tex]
- 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:
[tex]\text{Molarity of KOH}=\frac{0}{0.050}=0M[/tex]
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.
