A student performed a serial dilution on a stock solution of 1.33 M NaOH. A 1.0 mL aliquot of the stock NaOH (ms) was added to 9 mL of water to make the first dilution (m1). Next, 1.0 mL of the m1 solution was added to 9 mL of water to make the second solution (m2). The processes was repeated for a total of 5 times. What is the final concentration of NaOH (m5)

Question

Chemistry

Punk

2 May, 14:22

A student performed a serial dilution on a stock solution of 1.33 M NaOH. A 1.0 mL aliquot of the stock NaOH (ms) was added to 9 mL of water to make the first dilution (m1). Next, 1.0 mL of the m1 solution was added to 9 mL of water to make the second solution (m2). The processes was repeated for a total of 5 times. What is the final concentration of NaOH (m5)

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Answers (1)

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Eve Duke · Answer 1

The correct answer is 1.33 x 10⁻⁵ M

Explanation:

The concentration of the stock solution is: C = 1.33 M

In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:

C₁ = 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M

The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:

C₂ = 0.133 M x 1 ml/10 ml = 0.133 M x 1/10 = 0.0133 M

Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:

Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10

= initial concentration x (1/10) ⁵

= 1.33 M x 1 x 10⁻⁵

= 1.33 x 10⁻⁵ M