Questions 16-21 pertain to the titration of 30 mL of 0.250 M...

Questions 16-21 pertain to the titration of 30 mL of 0.250 M diprotic carbonic acid ($H_2CO_3$) solution with 0.750 M KOH. (p$K_{a1}$ = 6.35 and p$K_{a2}$ = 10.33.) (18) What will be the solution pH when a total of 8.00 mL of titrant have been added to the solution? (A) 4.13 (B) 10.93 (C) 5.75 (D) 3.53 (E) 6.95 (just used moles HA$^-$ in HH) (used p$K_{a2}$) (HH is upside down) (just used moles $H_2A$ in HH)

Transcript

00:01 The reaction between a weak acid and a strong base produces a basic solution.

00:13 Consider formic acid which is a weak acid reacting with sodium hydroxide or potassium hydroxide which is a strong base in order to form potassium formate and water according to this equation.

00:31 Now the salt potassium formate dissociates into potassium ion and formate ion when the formate ion which is the conjugate base of the weak acid will hydrolyze in water in order to form a basic solution according to this equilibrium equation.

00:56 The presence of hydroxide ion makes the solution basic.

01:01 We can solve for the poh of this basic solution using the following derived formula negative log of the square root of the pp of the conjugate base of the acid times the concentration of the salt.

01:20 So in this case we have a solution that is formed or that is we have a equation rather between 25 ml of 0 .2 molar formic acid with 0 .1 molar of koh.

01:43 So what we will do first is we're going to compute for the ml of koh.

01:49 So first we find the millimoles of the formic acid by multiplying its molarity by the volume in ml.

01:59 So we have molarity of 0 .2 volume in ml of 25 with this we get 5 millimoles.

02:10 So since at equilibrium at equivalence point the millimoles of the acid equals the millimoles of the base so the millimoles of the base koh is also equal to the millimoles of the acid and that's that means it's equal to 5.

02:25 So that the ml of koh is equal to its millimoles divided by its molarity...

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