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 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.

Answer 1

a) Answer :-
pH = 2.17

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

Given :-

[C₈H₆O₂] = 0.15 M

​​​​​p^Ka₁ = 3.51 and p^{Ka₂ = 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 – 18900'0 -01 XIST (0) (1 + 18900'0)

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 p^H using these two values of H₃O⁺ concentration which also gives same value of p^H

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