Respuesta :
Answer: The volume of barium hydroxide needed is 20.64 mL
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of KHP = 2.75 g
Molar mass of KHP = 204.22 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of KHP}=\frac{2.75g}{204.22g/mol}=0.0135mol[/tex]
The chemical reaction for the reaction of KHP and barium hydroxide follows:
[tex]2KHC_8H_4O_4(aq.)+Ba(OH)_2\rightarrow Ba(KC_8H_4O_4)_2(aq.)+2H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of KHP reacts with 1 mole of barium hydroxide
So, 0.0135 moles of KHP will react with = [tex]\frac{1}{2}\times 0.0135=0.00675mol[/tex] of barium hydroxide.
To calculate the molarity of barium hydroxide, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Moles of barium hydroxide = 0.00675 moles
Molarity of solution = 0.327 M
Putting values in above equation, we get:
[tex]0.327=\frac{0.00675mol\times 1000}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=\frac{0.00675\times 1000}{0.327}=20.64mL[/tex]
Hence, the volume of barium hydroxide needed is 20.64 mL
