A student prepared a stock solution by dissolving 2.50 g of KOH in enough water to make 150. mL of solution. She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution. What is the concentration of KOH for the final solution?

The correct answer is 6.9 x 10⁻⁵ M (0.0037 g/L)

Explanation:

The initial concentration of KOH solution can be calculated as follows:

mass= 2.50 g

volume = 150.0 ml

molecular weight of KOH= 39 g/mol + 16 g/mol + 1 g/mol= 56 g/mol

Concentration in g/L = mass/volume = 2.50 g/150.0 ml= 0.016 g/L

Concentration in M = 0.016 g/L x 1 mol/56 g = 3.0 x 10⁻⁴ M

When the student performed the dilution, she took 15.0 ml of initial solution and added water until reaching 65.0 ml. She performed a dilution in a factor of 15 ml/65.0  ml = 3/13. We calculate the final concentration of KOH as follows:

Final concentration = Initial concentration x dilution factor

                                = 3.0 x 10⁻⁴ M x 15 ml/65 ml

                                = 6.9 x 10⁻⁵ M

In g/L is the same procedure:

Final concentration = 0.016 g/L x 15 ml /65 ml = 4.6 x 10⁻³ g/L = 0.0037 g/L

it is a solid

explanation: