Respuesta :
Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 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 NaOH:
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol[/tex]
For sulfuric acid:
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
[tex]0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid = [tex](4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sodium sulfate
- For sodium sulfate:
Moles of sodium sulfate = [tex]1.462\times 10^{-3}moles[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M[/tex]
- For sulfuric acid:
Moles of excess sulfuric acid = [tex]3.498\times 10^{-3}mol[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M[/tex]
- For NaOH:
Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of NaOH}=\frac{0}{0.050}=0M[/tex]
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
