MarivelisseR1676 MarivelisseR1676
  • 20-02-2019
  • Chemistry
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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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superman1987
superman1987 superman1987
  • 01-03-2019

Answer:

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.

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