Question

"A student prepares a solution by dissolving 1.66 g of solid KOH in enough water to make 500.0 mL of solution. Calculate the molarity of K+ ions in this solution.
A 35.00 mL sample of this KOH solution is added to a 1000 mL volumetric flask, and water is added to the mark. What is the new molarity of K+ ions in this solution?"

Answer 1

Answer:

For 1: The molarity of $K^+\text{ ions}$ in this solution is 0.0592 M

For 2: The new molarity of $K^+\text{ ions}$ in this solution is $2.07\times 10^{-3}M$

Explanation:

• For 1:

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

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

Given mass of KOH = 1.66 g

Molar mass of KOH = 56.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{1.66\times 1000}{56.1g/mol\times 500.0}\\\\\text{Molarity of solution}=0.0592M$

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

So, molarity of $K^+\text{ ions}=0.0592M$

Hence, the molarity of $K^+\text{ ions}$ in this solution is 0.0592 M

• For 2:

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

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated KOH solution  having $K^+\text{ ions}$

$M_2\text{ and }V_2$ are the molarity and volume of diluted KOH solution  having $K^+\text{ ions}$

We are given:

$M_1=0.0592M\\V_1=35.00mL\\M_2=?M\\V_2=1000mL$

Putting values in above equation, we get:

$0.0592\times 35.00=M_2\times 1000\\\\M_2=\frac{0.0592\times 35.0}{1000}=2.07\times 10^{-3}M$

Hence, the new molarity of $K^+\text{ ions}$ in this solution is $2.07\times 10^{-3}M$