Question

In the titration f 25mL of 0.1000M maleic acid, HOOC-CH=CH-COOH with 0.1000 M NaOH. Symbolizing the acid as H2M, we can write the two dissociation equilibra as:

H2M+H2O <-> HM^- +H3O^+ K(a1)=1.3x10^-2

HM^- +H2O <-> M^2- +H3O^+ K(a2)=5.9x10^-7

1) Determine the [H3O^+], total volume, Cacid, [Na], and [OH] when 5mL of NaOH were added

2) Determine the [H3O^+], total volume, Cacid, [Na], and [OH] when 25.5mL of NaOH were added

PLEASE HELP SOLVE! NEED SOON!

Answer 1

1 ) 5 mL NaOH added

millimoles of acid H2A = 25 x 0.1 = 2.5

millimoles of NaOH = 5 x 0.1 = 0.5

total volume = 25 + 5 = 30 mL

H2A + OH- ----------------------> HA- + H2O

2.5 0.5 0 0

2.0 0 0.5

[C] acid = millimoles remains / total volume

= 2.0 / 30

= 0.067 M

[Na] = 0.5 / 30

= 0.0167 M

Ka1 = 1.3 x 10^-2

pKa1 = 1.89

pH = pKa1 + log [HA-/H2A]

pH = 1.89 + log (0.5 /2)

pH = 1.29

[H3O+] = 10^-1.29

= 0.051 M

[OH-] = Kw / [H3O+]

= 1.0 x 10^-14 / 0.051

= 1.95 x 10^-13 M

2 )

millimoles of NaOH = 0.1 x 25.5 = 2.55

H2A + 2NaOH ------------------> Na2A + H2O

2.5 2.55 0 0

0.0 0.05 2.5 0

total volume = 25 + 25.5 = 50.5 mL

[C] = 0.0 M

[Na+] = (0.05 + 2 x 2.5 ) / 50.5

= 0.1 M