Question

A chemist dissolves 697. mg of pure potassium hydroxide in enough water to make up 160. mL of solution. Calculate the pH of the solution.
(The temperature of the solution is 25 °C.)
Be sure your answer has the correct number of significant digits.

Answer 1

Answer: The pH of the solution is 12.9

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

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:

$\text{Molarity of KOH solution}=\frac{0.697\times 1000}{56\times 160}\\\\\text{Molarity of KOH solution}=0.0778M$

1 mole of KOH produces 1 mole of $K^+$ ions and 1 mole of $OH^-$ ions

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

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.0778M$

Putting values in above equation, we get:

$pOH=-\log (0.0778)=1.11$

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

$pH+pOH=14\\pH=14-1.11=12.89$

Hence, the pH of the solution is 12.9