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The volume of a sample of pure HCl gas was 189 mL at 25°C and 108 mmHg. It was completely dissolved in about 60 mL of water and titrated with an NaOH solution; 15.7 mL of the NaOH solution were required to neutralize the HCl. Calculate the molarity of the NaOH solution.

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Answer:

M(NaOH) = 0.07 M

Explanation:

In a neutralization such as this the moles of the acid are the same as the base because you are titrating it until that occurs.

The moles of HCl can be obtained with the gas information using PV = nRT keep in mind to use consistent units.

[tex]PV= nRT \\\frac{108}{760} \frac{189}{1000}=n(0.0821)(25+273) \\n= 0.00110\ moles[/tex]

The molarity of NaOH:

[tex]M = \frac{n}{V}[/tex]

Because the moles of HCl are the same as NaOH

[tex]M = \frac{0.00110}{15.7/1000} =0.070 M[/tex]

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