A chemist dissolves 716.mg of pure potassium hydroxide in enough water to make up 130.mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 degree C.) Be sure your answer has the correct number of significant digits.

[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 = 716. mg = 0.716 g (Conversion factor: 1 g = 1000 mg)

Molar mass of KOH = 56 g/mol

Volume of solution = 130 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{0.716\times 1000}{56g/mol\times 130}\\\\\text{Molarity of solution}=0.098M[/tex]

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions

To calculate hydroxide ion concentration of the solution, we use the equation:

[tex]pOH=-\log[OH^-][/tex]

We are given:

[tex[[OH^-]=0.098M[/tex]

Putting values in above equation, we get:

[tex]pOH=-\log(0.098)\\\\pOH=1.00[/tex]

To calculate the pH of the solution, we use the equation:

pOH + pH = 14

So, pH = 14 - 1.00 = 13.0

Hence, the pH of the solution is 13.0