Respuesta :
Answer : The pH of the solution is, 13.2
Explanation :
First we have to calculate the mole of [tex]Ba(OH)_2[/tex] and [tex]KOH[/tex].
[tex]\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }Ba(OH)_2=0.073M\times 0.047L=0.00343mol[/tex]
and,
[tex]\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }KOH=0.20M\times 0.054L=0.0108mol[/tex]
Now we have to calculate the moles of hydroxide ion.
As, 1 mole of [tex]Ba(OH)_2[/tex] dissociates to give 1 mole of barium ion and 2 mole of hydroxide ion.
So, 0.00343 mole of [tex]Ba(OH)_2[/tex] dissociates to give 0.00343 mole of barium ion and (2×0.00343=0.00686) mole of hydroxide ion.
and,
As, 1 mole of [tex]KOH[/tex] dissociates to give 1 mole of potassium ion and 1 mole of hydroxide ion.
So, 0.0108 mole of [tex]KOH[/tex] dissociates to give 0.0108 mole of potassium ion and 0.0108 mole of hydroxide ion.
Now we have to calculate the total moles of hydroxide ion and volume of solution.
Total moles of chloride ion = 0.00686 + 0.0108 = 0.01766 mol
Total volume of solution = 47 mL + 54 mL = 101 mL = 0.101 L
Now we have to calculate the molarity of chloride ion in the final solution.
[tex]\text{Molarity}=\frac{\text{Total moles}}{\text{Total volume}}[/tex]
[tex]\text{Molarity}=\frac{0.01766mol}{0.101L}=0.175M[/tex]
Now we have to calculate the value of pOH of the solution.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (0.175)[/tex]
[tex]pOH=0.757[/tex]
Now we have to calculate the value of pH of the solution.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.757\\\\pH=13.2[/tex]
Therefore, the pH of the solution is, 13.2
The pH of the solution is 13.2. Firstly we need to calculate the molarity of the solution.
Calculation of pH:
First, we have to calculate the mole of Ba(OH)₂ and KOH.
Moles of Ba(OH)₂ = Molarity * Volume of solution
Moles of Ba(OH)₂ = 0.073 * 0.047 L=0.00343 mol
and,
Moles of KOH = Molarity * Volume of solution
Moles of KOH = 0.20 M * 0.054 L = 0.0108 mol
Now we have to calculate the moles of hydroxide ion.
As, 1 mole of Ba(OH)₂ dissociates to give 1 mole of barium ion and 2 mole of hydroxide ion.
So, 0.00343 mole of Ba(OH)₂ dissociates to give 0.00343 mole of barium ion and (2×0.00343=0.00686) mole of hydroxide ion.
and,
As, 1 mole of KOH dissociates to give 1 mole of potassium ion and 1 mole of hydroxide ion.
So, 0.0108 mole of KOH dissociates to give 0.0108 mole of potassium ion and 0.0108 mole of hydroxide ion.
Calculation of total moles of hydroxide ion and the volume of solution.
Total moles of chloride ion = 0.00686 + 0.0108 = 0.01766 mol
Total volume of solution = 47 mL + 54 mL = 101 mL = 0.101 L
Calcluation of Molarity:
Molarity = Number of moles / Volume of solution
Molarity = 0.01766 moles / 0.101 L = 0.175 M
Calculation of pOH of the solution:
pOH = -log[OH-]
pOH = -log[0.175]
pOH = 0.757
Calcluation of pH of the solution:
pH+ pOH =14
pH + 0.757 = 14
pH = 13.2
Therefore, the pH of the solution is 13.2
Find more information about pH here:
brainly.com/question/13557815
