The volume of a sample of pure HCl gas was 221 mL at 20°C and 111 mmHg. It was completely dissolved in about 50 mL of water and titrated with an NaOH solution; 18.7 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

Answer:

[tex]molarity =6.9\times 10^{-3}\ M\\[/tex]

Explanation:

We know that , the reaction of HCl and NaOH is given as follows

[tex]NaOH+HCl=NaCl +H_2O[/tex]

Given that

Pressure = 111 mm Hg

[tex]P=111\times 13.6\times 10^{-3}\times 9.81\times 1000=14.809\ kPa[/tex]

Temperature = 20°C

T=20+273=293 K

Volume= 221 m L

V=0.221 L

Number of moles of HCl is given as follows

[tex]n=\dfrac{P\times V}{R\times T}\\n=\dfrac{0.148\times 0.221}{0.821\times 293}=1.3\times 10^{-4}\ moles[/tex]

From the above reaction we can say that

Number of moles of HCl=Number of moles of NaOH

Volume of NaoH is given as follows

V=18.7 = 0.0187 L

Therefore molarity

[tex]molarity =\dfrac{n}{V_{NaOH}}\\molarity =\dfrac{1.3\times 10^{-4}}{0.0187}=6.9\times 10^{-3}\ M\\molarity =6.9\times 10^{-3}\ M\\[/tex]

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