Question

Phathalic acid, H2C8H4O4, is a diprotic acid and has a concentration of 2.9 M. Ka1 = 0.0012 and Ka2 = 3.9 x 10-6. Estimate the concentration of [HC8H4O4-] and [C8H4O4-2] at equilibrium. Determine Kb1 and Kb2. Determine the pH of the solution at each proton dissociation. 11. Calculate the molar solubility of calcium hydroxide (Ksp = 4.8 x 10^6

I NEED TO KNOW HOW DID THEY GET THIS ANSWER PLEASE 0.0012 = x2/2.9 x = 0.059 M equilibrium concentration of HC8H4O4- = 0.059 M

Answer 1

First, write the two equatiosn of dissociation:

H₂C₈H₄O₄ ------------> HC₈H₄O₄^- + H⁺   Ka1 = 0.0012

HC₈H₄O₄^- ------------> C₈H₄O₄^2- + H⁺   Ka2 = 3.9x10^-6

Now, let's calculate the first concentration:

H₂C₈H₄O₄ ------------> HC₈H₄O₄^- + H⁺

i. 2.9 0 0

e. 2.9-x x x

0.0012 = x² / 2.9-x Ka is a low value, so 2.9-x can be neglected so 2.9-x = 2.9

0.0012 = x² / 2.9

x = (0.0012 * 2.9)^1/2

x = 0.059 M

Now for the second reaction:

HC₈H₄O₄^- ------------> C₈H₄O₄^2- + H⁺

i. 0.059 0 0

e. 0.059-x x x

3.9x10^-6 = x² / 0.059-x

x = (0.059 * 3.19x10^-6)^1/2

x = 4.33x10^-4 M

For Kb: Kb = Kw / Ka

Kb1 = 1x10^-14 / 0.0012 = 8.33x10^-12

Kb2 = 1x10^-14 / 3.9x10^-6 = 2.56x10^-9

Finally the pH:

pH = -log(H⁺)

[H⁺] is the same concentration of the ions calculated above (see reaction) so:

pH1 = -log(0.059) = 1.23

pH2 = -log(4.33x10^-4) = 3.36

For the second question:

Ca(OH)₂ ---------> Ca²⁺ + 2OH^-

i. s 0 0

e. 0 s 2s

Ksp = [Ca²⁺] [OH^-]²

4.8x10^-6 = (s) * (2s)²

4.8x10^-6 = 4s³

s = (4.8x10^-6 / 4 )^1/3

s = 0.011 M

Hope this helps. Any doubt or question, you may tell me in a comment.