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3 years ago

35mL of acid with an unknown concentration is titrated to completion using 63mL of 3.0 MNaOH.

Chemistry

1 answer:

siniylev [52]3 years ago

4 0

Answer:

5.4 M.

Explanation:

At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.

M of acid = ??? M, V of acid = 35.0 mL.

M of NaOH = 3.0 M, V of NaOH = 63.0 mL.

∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.

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