If 18.9 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?

Question

Tabetha

Chemistry

26 September, 08:24

If 18.9 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?

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Answers (1)

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Magdalena Hess · Answer 1

3.024 M.

Explanation:

Based on at the neutralization, the no. of millimoles of acid (HCl) is equal to the no. of millimoles of the base (ammonia).

(MV) HCl = (MV) ammonia

M of HCl = 0.80 M & V of HCl = 18.9 mL.

M of ammonia = ? M & V of ammonia = 5.0 mL.

∴ M of ammonia = (MV) HCl / V of ammonia = (0.80 M) (18.9 mL) / (5.0 mL) = 3.024 M.