Question

I) You have an HCl solution of pH 3.13.

(a) The concentration of this solution is ______________ M

(b) If you increased the HCl concentration exactly 2 fold, the pH would be ______________

(c) If you decrease the pH to 2.13, the new HCl concentration would be ______________

II) To 20.000 mL of distilled water in a 50 mL beaker, you add exactly 13 drops of 0.100 M HCl and swirl to mix. If exactly 18 drops of HCl are required to make 1.000 mL

Calculate: Molarity of the diluted HCl ____________

M The pH of the diluted HCl ____________

Answer 1

Solution:- (I) (a) HCl is a strong acid so it dissociate completely to give H⁺. The H⁺ ion concentration would be equal to the HCl concentration.

pH = - log[H⁺]

[H⁺] = 10^-pH = 10^-3.13 = 0.000741 M

So, the HCl concentration is 0.000741 M or 7.41 x 10^-4 M.

(b) If the concentration is increased 2 fold, then new concentration would be 2 x 7.41 x 10^-4 M = 1.482 x 10^-3 M.

pH = - log(1.482 x 10^-3)

pH = 2.83

(c)  If the pH is decreased to 2.13 then the new concentration could be calculated as..

[H⁺] = 10^-2.13 = 0.00741 M

So, the new concentration of HCl solution would be 0.00741 M or 7.41 x 10^-3 M.

(II) Volume of 18 drops of HCl is 1.000 mL, then we could calculate the volume of 13 drops of HCl.

13 drops x (1.000 mL/18 drops) = 0.7222 mL

20.000 mL of distilled water is used to dilute the HCl solution as the 13 drops of the acid are added to this 20.000 mL of distilled water.

We would use the dilution equation to calculate the concentration of diluted solution.

M₁V₁ = M₂V₂

M₂ = M₁V₁/V₂

Initial volume, V₁ = 0.7222 mL

Final volume, V₂ = 0.7222 mL + 20.000 mL = 20.722 mL

M₁ = 0.100 M

M₂ = ?

M₂ = 0.7222 mL x 0.100 M/20.722 mL = 0.00349 M

pH = - log 0.00349

pH = 2.46