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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.

Respuesta :

joseaaronlara

Answer:

The answer to your question is: Molarity = 0.078

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

           

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