Question

A student is given 3 beakers:

Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak 
monoprotic acid ,HX, in enough water to produce 1 liter of solution. 
The empirical formula of HX is CH
O. The solution contains 3 drops 
of phenolphthalein.

Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

Beaker 3 - 50.0 ml of 0.250M KOH

The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.

Answer 1

Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

X- + H2O    HX + OH-
Initial 0.07 0 0
Change -x x x
Final 0.07 - x x x

pOH = 14 - pH

= 14 - 8.8 = 5.2

[OH-] = x = 10^-5.2 = 6.31*10^-6

Kb of X- = x*x / (0.07 - x)

= (6.31*10^-6)² / (0.07 - 6.31*10^-6)

Kb = 5.68*10^-10

Ka = 10^-14 / 5.68*10^-10

= 1.76*10^-5

pKa = - log Ka

= - log (1.76*10^-5)

= 10.94

From Beaker #1:

Mass of acid = 6.00 grams

Molar mass of acid (CH2O) = 30 g/mole

Moles of acid = 6 / 30 = 0.2

Volume of solution = 1 L

[HX] = 0.2 M

Millimoles of HX = 0.2 * 50 = 10 mmols

50.0 ml of 0.250M KOH

20.0 ml of the beaker 3 solution is used.

Milimoles of KOH = 20 * 0.250 = 5.0 mmols

HX + KOH    KX + H2O
Initial 10 5 0 0
Final 5 0 5 5

Using Henderson-Hasselbalch equation:

pH = pKa + log [base] / [acid]

= 10.94 + log (5 / 5)

pH = 10.94

The Ka value = 1.76*10^-5 corresponds to acetic acid.

So X will be acetate ion (CH3COO-)