Question

If 35 mL of a .5 m hydrochloric acid solution is required to neutralize 15 mL of a sodium hydroxide solution of unknown concentration, what is the molarity of the sodium hydroxide solution

Answer 1

Answer: The molarity of the sodium hydroxide solution is 1.16 M.

Explanation:

Given: $V_{1}$ = 35 mL,    $M_{1}$ = 0.5 M

$V_{2}$ = 15 mL,           $M_{2}$ = ?

Formula used to calculate the molarity of the sodium hydroxide solution is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.5 M \times 35 mL = M_{2} \times 15 mL\\M_{2} = \frac{0.5 M \times 35 mL}{15 mL}\\= 1.16 M$

Thus, we can conclude that the molarity of the sodium hydroxide solution is 1.16 M.