Question

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 1

`molarity =6.9* 10^(-3)\ M\\`

Step-by-step explanation:

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

`NaOH+HCl=NaCl +H_2O`

Given that

Pressure = 111 mm Hg

`P=111* 13.6* 10^(-3)* 9.81* 1000=14.809\ kPa`

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

`n=(P* V)/(R* T)\\n=(0.148* 0.221)/(0.821* 293)=1.3* 10^(-4)\ moles`

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

`molarity =(n)/(V_(NaOH))\\molarity =(1.3* 10^(-4))/(0.0187)=6.9* 10^(-3)\ M\\molarity =6.9* 10^(-3)\ M\\`