Question

The volume of a sample of pure HCl gas was 205 mL at 27°C and 141 mmHg. It was completely dissolved in about 70 mL of water and titrated with an NaOH solution; 24.3 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

Answer 1

The answer to your question is: Molarity = 0.078

Step-by-step explanation:

Data

HCl

V = 250 ml

T = 27°C = 300 °K

P = 141 mmHg = 0.185 atm

V2 = 70 ml

NaOH

V = 24.3 ml

Molarity NaOH = ?

Process

1.- Calculate the number of moles of HCl

PV = nRT

n = PV / RT

R = 0.082 atm l / mol K

n = (0.185)(0.25) / (0.082)(300)

n = 0.046 / 24.6

n = 0.0019 moles

2.- Calculate molarity of HCl

Molarity = moles / volume

Molarity = 0.0019 / 0.070

Molarity = 0.027

3.- Write the balanced equation

HCl + NaOH ⇒ H₂O + NaCl

Here, we observe that the proportion HCl to NaOH is 1:1 .

Then 0.0019 moles of HCl reacts with 0.0019 moles of NaOH.

4.- Calculate the molarity of NaOH.

Molarity = 0.0019 / 0.0243

Molarity = 0.078