If I add 375 of water to 125 mL of a 0.90 M NaOH solution, what will the molarity of the diluted solution be?

0.225M

Explanation:

Step 1:

Data obtained from the question. This includes:

Volume of the stock solution (V1) = 125mL

Molarity of the stock solution (M1) = 0.90 M

Volume of the diluted solution (V2) = 125 + 375 = 500mL

Molarity of the diluted solution (M2) =..?

Step 2:

Determination of the molarity of the diluted solution.

This is obtained by using the dilution formula as follow:

M1V1 = M2V2

0.9 x 125 = M2 x 500

Divide both side by 500

M2 = (0.9 x 125) /500

M2 = 0.225M

Therefore, the molarity of the diluted solution is 0.225M

The molarity of the diluted solution of NaOH is 0.225 M NaOH

Explanation:

Here, we have, initial concentration of the solution = 0.90 M NaOH

Volume of initial solution = 125 mL

Volume of added water = 375 mL

Total volume of diluted solution = 375 mL + 125 mL = 500 mL

Number of moles in the diluted solution = 0.9 × 0.125 = 0.1125 moles of NaOH

Derived from the following relation;

0.9 M NaOH indicates that 1000 mL contains 0.9 moles of NaOH

Therefore, 1 mL contains 0.9/1000 moles of NaOH and 125 mL contains 0.9/1000×125 or 0.9 × 0.125 moles of NaOH

Therefore, the diluted 500 mL solution contains 0.1125 moles of NaOH. Therefore, 1000 mL of the diluted solution will contain of NaOH or 0.225 moles of NaOH

Hence the molarity of the diluted solution of NaOH, which is the number of moles per 1000 mL is 0.225 M.

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