The volume of a sample of pure HCl gas was 161 mL at 26°C and 139 mmHg. It was completely dissolved in about 60 mL of water and titrated with an NaOH solution; 27.7 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

Question

The volume of a sample of pure HCl gas was 161 mL at 26°C and 139 mmHg. It was completely dissolved in about 60 mL of water and titrated with an NaOH solution; 27.7 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

Answer 1

First, we need to calculate the number of moles of HCl in the original gas sample:

PV = nRT

n = PV/RT

n = (139 mmHg)(0.161 L)/(0.0821 L·atm/mol·K)(299 K)

n = 0.00757 mol

This is the number of moles of HCl in the original gas sample. Since the HCl is completely dissolved in water and neutralized by NaOH, the number of moles of NaOH required to neutralize the HCl is also 0.00757 mol. We can use this information to calculate the molarity of the NaOH solution:

mol NaOH = 0.00757 mol

Molarity = mol/L = 0.00757 mol / 0.0277 L = 0.273 M

Therefore, the molarity of the NaOH solution is 0.273 M.