Question

Ka for hypochlorous acid, HClO, is 3.0 x 10^-8. Calculate the pH after 10.0, 20.0, 30.0, and 40.0 ml of 0.100M NaOH have been added to 40.0ml of 0.100M HClO. Expert Answer GoldenApple6699 GoldenApple6699 answered this Was this answer helpful? 8 0 766 answers At any point between 0 and 40 mL of NaOH added, you will have a solution containing both HClO and ClO- ... that's a buffer system (a weak acid and its conjugate base) and the pH is governed by the Henderson-Hasselbalch equation: pH = pKa + log [ClO-]/[HClO] which is equivalent to pH = pKa + log (moles ClO- / moles HClO) Initial moles HClO = M HClO x L HClO = (0.100)(0.0400) = 0.00400 moles HClO initially At 10.0 mL NaOH: moles NaOH added = M NaOH x L NaOH = (0.100)(0.0100) = 0.00100 moles NaOH and HClO react in a 1:1 mole ratio. So the 0.00100 moles will react with 0.00100 moles of HClO to produce 0.00100 moles of NaClO (ClO- ion). There are 0.00400 - 0.00100 = 0.00300 moles of HClO left. pH = pKa + log (moles ClO- / moles HClO) pH = -log (3.0 x 10^-8) + log (0.00100 / 0.00300) = 7.52 - 0.48 = 7.04 At 20.0 mL NaOH: moles NaOH = 0.00200 which react with 0.00200 moles HClO forming 0.00200 moles ClO-. There are 0.00400 - 0.00200 = 0.00200 moles HClO left. pH = 7.52 + log (0.00200 / 0.00200) = 7.52 + 0 = 7.52 At 30.0 mL NaOH: 0.00300 moles of NaOH react with 0.00300 moles of HClO to produce 0.00300 moles of Cl)-. There are 0.00400 - 0.00300 = 0.00100 moles of HClO left. pH = 7.52 + log (0.00300 / 0.00100) = 7.52 + 0.48 = 8.00 At 40.0 mL NaOH: this is the equivalence point where moles NaOH (0.00400) = moles HClO (0.00400). The only thing left is the 0.00400 moles of ClO- produced; since the total solution volume is now 80 mL, M ClO- = moles ClO- / L solution = 0.00400 / 0.0800 = 0.0500 M ClO-. Since it was produced from a strong base (NaOH) and a weak acid (HClO), the salt will give a basic solution in water. The hydrolysis reaction is (note that it produces OH- like a base should): Set up an ICE chart. ClO- will produce small and equal amounts of HClO and OH-. Molarity . . . . .ClO- + H2O <==> HClO + OH- Initial . . . . . . 0.0500 . . . . . . . . . . .0 . . . .0 Change . . . . . .-x . . . . . . . . . . . . .x . . . .x at Equilibrium .0.0500-x . . . . . . . . .x . . . .x Since ClO- is the conjugate base of HClO, Kb ClO- = Kw / Ka = (1 x 10^-14 / 3.0 x 10^-8) = 3.3 x 10^-7 Kb = [HClO][OH-] / [HClO] = (x)(x) / (0.0500-x) = 3.3 x 10^-7 Since Kb is small, the -x term after 0.0500 will be essentially negligible, so this simplifies to x^2 / 0.0500 = 3.3 x 10^-7 x^2 = 1.7 x 10^-8 x = 1.3 x 10^-4 = [OH-] pOH = -log [OH-] = -log (1.3 x 10^-4) = 3.89 pH = 14.00 - pOH = 14.00 - 3.89 = 10.11 Comment

I am clueless from this answer!!!!! Can anyone break it down a little further? I appreciate the assistance. Please note: I am stupid and need all the steps!

Answer 1

Ka of HClO = 3.0 x 10^-8.

pKa = - log Ka = - log (3.0 x 10^-8)

        = 7.52

millimoles of HClO = 40. 0 x 0.100 = 4

a) pH after addition of 10.0 mL NaOH :

millimoles of NaOH = 10 x 0.1 = 1

HClO   +   NaOH    ------------> NaClO +   H2O

   4               1                             0                 0

3                  0                             1

pH = pKa + log [salt / acid]

     = 7.52 + log (1 / 3)

pH = 7.22

b) 20.0 mL

mmoles of NaOH = 20 x 0.1 = 2

HClO   +   NaOH    ------------> NaClO +   H2O

   4               2                             0                 0

   2                  0                           2

pH = pKa + log [salt / acid]

     = 7.52 + log (2 / 2)

pH = 7.52

c) 30.0 mL NaOH :

mmoles of NaOH = 30 x 0.1 = 3

HClO   +   NaOH    ------------> NaClO +   H2O

   4               3                             0                 0

1                  0                            3

pH = pKa + log [salt / acid]

     = 7.52 + log (3 / 1)

pH = 8.00

d) 40.0 ml NaOH

mmoles of NaOH = 40 x 0.1 = 4

HClO   +   NaOH    ------------> NaClO +   H2O

   4               4                             0                 0

0                  0                            4

here salt only remains.

salt concentration = mmoles / total volume

                             = 4 / 40 + 40 = 0.05 M

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (7.52 + log 0.05)

pH = 10.11