Question

Terephthalic (C8H6O4) is a diprotic acid with pka's 3.51 and 4.82. a. What is the pH of a 0.15M solution of this aci...

Terephthalic (C8H6O4) is a diprotic acid with pka's 3.51 and 4.82.

a. What is the pH of a 0.15M solution of this acid? [x1] ( write answer to hundredths place)

b. What is the concentration of C8H4O4-2 in a 0.15M solution f this acid? [x2] ( write answer to three significant figures and use "E" to write exponent)

I think (a) pH is 2.18 but I need concentration.

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Answer #1

a) Answer :- pH = 2.17

C8H04(aq) + H20(1) = H30 Tag) + C3H502 (aq) ...(A)

Given :-

[C8H6O2] = 0.15 M

​​​​​pKa1 = 3.51 and pKa2 = 4.82

We know that,

pKa = -logk,    ie. (oyd - )boud = y

therefore,

Kal = antilog(-pKal) = antilog(-3.51) = 3.09 x 10-4

K12 = antilog(-pKq2) = antilog(-4.82) = 1.51 x 10-5

Now, for equilibrium (A) the ICE table is

CH604(aq) + H2O1) \rightleftharpoons H30 taq + C8H50(aq)
initial 0.15 - 0 0
change 0.15 - x +x +x
equilibrium 0.15 - x x x

Now,

[H3O+][C8H502] C3H604]

i.e -= 3.09 x 10-4 0.15 -

but 0.15 >>x ie.   

= 3.09 x 10-4

c? = 0.15 X 3.09 x 10-4 = 4.64 x 10-5

I = 6.81 x 10-3M = 0.00681M

i.e [H3O+] = [H+] = 0.00681M

Now,

pH = -log[+] = -log(0.00681) = -(-2.17)  

pH = 2.17

-------------------

b) Answer :- CH20-1 = 1.51E-5M

Again CH:02 (aq) + H2O) = H30 Tag) +C8H4O2 (aq)

The ICE table is

C8H50(aq) + H2O1) \rightleftharpoons H30 taq + C8H40% (aq)
I 0.00681 - 0.00681 0
C 0.00681 - x - 0.00681 + x + x
E 0.00681 - x - 0.00681 + x x

Now,

[EoʻHo] [LOH$]+0H

I – 189000 -01 XIST (0) (1 + 189000)

but, (0.00681+c)×(0.00681-x)

therefore,

r = 1.51 x 10-5M

ie. C8H402-) = = 1.51 x 10-5M

ie. C8H402-1= r = 1.51E-5M

------------------

Note:-

we can calculate pH using these two values of H3O+ concentration which also gives same value of pH

ie. [H3O+] = 6.81 x 10-3 +1.51 x 10-5 = 6.8251 x 10-3

ie. pH = -log(6.8251 x 10-3) = -(-2.17) = 2.17

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