Question

Ka for hypochlorous acid. HCIO, is 3.0x108. Calculate the pH after the stepwise addition of 0.100 M NaOH (from 0 ml to 50 mL, in I mL increments) to 40.0 ml of 0.100 M HCIO. Plot your results in Excel and on the graph, identify the pH at the equivalence point and at half of the equivalence point.

Answer 1

(i) The volume of NaOH solution = 0 mL

The concentration of HClO = [HClO] = 0.1 M

The Ka of HClO = 3*10^-8

i.e. pKa = -LogKa = -Log(3*10^-8) = 7.523

Now, pH = 1/2 (pKa - Log[HClO])

= 1/2 (7.523 - Log 0.1)

= 4.26

(ii) The volume of NaOH solution = 1 mL

i.e. The no. of mmol of NaOH (n_NaOH) = 1 mL * 0.1 mmol/mL = 0.1 mmol

The no. of mmol of HClO (n_HClO) = 40 mL * 0.1 mmol/mL = 4 mmol

Here, 0.1 mmol of NaOH reacts with 0.1 mmol of HClO to form 0.1 mmol of NaClO, the remaining n_HClO = 4-0.1 = 3.9

According to Henderson-Hasselbulch equaiton: pH = pKa + Log(n_NaOH/n_HClO)

i.e. pH = 7.523 + Log(0.1/3.9)

i.e. pH = 5.93

(iii) The volume of NaOH solution = 2 mL

n_NaOH = 0.2 mmol and n_HClO = 3.8 mmol

Now, pH = 7.523 + Log(0.2/3.8)

i.e. pH = 6.24

(iv) The volume of NaOH solution = 3 mL

n_NaOH = 0.3 mmol and n_HClO = 3.7 mmol

Now, pH = 7.523 + Log(0.3/3.7)

i.e. pH = 6.43

(v) The volume of NaOH solution = 20 mL

n_NaOH = 2 mmol and n_HClO = 2 mmol

Now, pH = 7.523 + Log(2/2)

i.e. pH = 7.52 which is the pH at half equivalence point

(vi) The volume of NaOH solution = 40 mL

n_NaOH = n_NaClO = 4 mmol and n_HClO = 0 mmol

Here, [NaClO] = 4 mmol/(40+40) mL = 0.05 M

Formula: pH = 7 + 1/2(pKa + Log[NaClO])

i.e. pH = 7 + 1/2 (7.523 + Log 0.05)

i.e. pH = 10.11 which is the pH at equivalence point

(vii) The volume of NaOH solution = 50 mL

n_NaOH = 5 mmol and n_NaClO = 4 mmol, i.e. the remaining n_NaOH = 5 - 4 = 1 mmol

Here, [NaOH] = [OH^-] = 1 mmol/(40+50) mL = 0.011 M

pOH = -Log[OH^-] = -Log(0.011)

i.e. pOH = 1.95

pH + pOH = 14

i.e. pH + 1.95 = 14

i.e. pH = 14 - 1.95 = 12.05

The plot of pH versus volume of 0.1 M NaOH solution can be drawn as follows.

14 12 10 10 20 30 40 50 Volume of 0.1 M NaOH solution