Answer: The pH of the solution is 12.9
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Given mass of KOH = 697 mg = 0.697 g (Conversion factor: 1 g = 1000 mg)
Molar mass of KOH = 56 g/mol
Volume of solution = 160 mL
Putting values in above equation, we get:
[tex]\text{Molarity of KOH solution}=\frac{0.697\times 1000}{56\times 160}\\\\\text{Molarity of KOH solution}=0.0778M[/tex]
1 mole of KOH produces 1 mole of [tex]K^+[/tex] ions and 1 mole of [tex]OH^-[/tex] ions
To calculate the pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex][OH^-]=0.0778M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log (0.0778)=1.11[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-1.11=12.89[/tex]
Hence, the pH of the solution is 12.9
