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lilavasa [31]
3 years ago
5

A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to a

chieve this pH, what was the original concentration of the sample of HCl? Enter your answer in normal notation (no exponents) and out to the 1/1000th place if the answer is less than 1. (answer in the form 0.00x; enter no units, although the number should correspond to Molarity)
Chemistry
1 answer:
MArishka [77]3 years ago
4 0

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

M = 0.04 mol / 1.2 L = 0.033 M

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