Question

On a lab exam, you have to find the concentrations of the monoprotic (one proton per molecule) acids HA and HB. You are given 42.6 mL of HA solution in one flask. A second flask contains 38.8 mL of HA, and you add enough HB solution to it to reach a final volume of 50.0 mL. You titrate the first HA solution with 87.3 mL of 0.0906 M NaOH and the mixture of HA and HB in the second flask with 96.4 mL of the NaOH solution. Calculate the molarity of the HA and HB solutions.

Answer 1

Molarity of NaOH (M₁) = 0.0906M

Volume of NaOH used by HA (V₁) = 87.3 mL

Volume of HA in flask 1 (V₂) = 42.6 mL

Let Molarity of HA in Flask 1 be M₂

From Molarity equation,

M₁V₁ =M₂V₂

0.0906M*87.3 mL = M₂* 42.6 mL

M₂ = 0.1857M

Thus, Molarity of HA solution is 0.1857 M

Now, Volume of HA in flask 2(V₂') = 38.8 mL

THus, Volume of NaOH used by HA(v₁') can be found as

M₁V_1' = M₂V₂'

0.0906 M * V₁' = 0.1857 M* 38.8 mL

V₁' = 79.5 mL

THus, Volume of NaOH used by HB (V₁") = 96.4- 79.5 = 16.9 mL

Volume ofHB in flask2 (V₂") = 50-38.8 = 11.2 mL

Now, M₁V₁" = M₂V₂"

0.0906M * 16.9 mL = M₂ * 11.2 mL

M₂ = 0.1367M

Thus, Molarity of HB is 0.1367M