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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 :

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

Explanation:

Given: [tex]V_{1}[/tex] = 35 mL,    [tex]M_{1}[/tex] = 0.5 M

[tex]V_{2}[/tex] = 15 mL,           [tex]M_{2}[/tex] = ?

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

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute the values into above formula as follows.

[tex]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[/tex]

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

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