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

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* 10^(-3)M`

Step-by-step explanation:

• For 1:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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}=(1.66* 1000)/(56.1g/mol* 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* 35.00=M_2* 1000\\\\M_2=(0.0592* 35.0)/(1000)=2.07* 10^(-3)M`

Hence, the new molarity of
`K^+\text{ ions}` in this solution is
`2.07* 10^(-3)M`