PLEASE HELP ASAP

Question

Calculate the pH of the following solutions,

0.0040 M benzoic acid (Ka= 6.28x10-5)

prepared by adding 100 mL of 0.0500 M NaOH to 50 mL of 0.100 M HOCl (Ka= 3.0x10-8)

0.012 M HClO4.

please write the solution with steps

Answer 1

Benzoic acid is a weak acid and we can represent it as HA

HA ==> H⁺ + A^-

Ka = 6.28x10^-5 = [H⁺][A^-]/[HA]

6.28x10^-5 = (x)(x)/0.004 - x

x² = 2.51x10^-7 - 6.28x10^-5x

x² + 6.28x10^-5x - 2.51x10^-7 = 0

x = 0.0004706 M = [H⁺] = 4.71x10^-4

pH = -log [H⁺] = - log 4.71x10^-4

pH = 3.33

moles NaOH present = 0.1 L x 0.0500 mol/L = 0.00500 moles

moles HOCl present = 0.05 L x 0.100 mol/L = 0.005 moles

NaOH + HOCl ==> NaOCl + H2O

0.005.....0.005.........0.............Initial

-0.005...-0.005......+0.005.......Change

0............0.............0.005.........Equilibrium

So, when the reaction is over, there is no NaOH and no HOCl remaining. All you have is NaOCl, which will react with water as follows:

OCl- + H2O ==> HOCl + OH- and we need to look up the Kb for OCl- or the Ka for HOCl. I find Ka HClO to be 3x10^-8 so from that we get a Kb of 1x10^-14/3x10^-8 = 3.33x10^-7

Kb = 3.33x10^-7 = [HOCl][OH-]/[OCl-] = (x)(x) / 0.005

x2 = 1.67x10^-9

x = 4.1x10^-5 = [OH-]

pOH = -log 4.1x10^-5 = 4.39

pH = 14 - 4.39

pH = 9.61