Use the method of successive approximations to determine the pH and concentratio

Question

Use the method of successive approximations to determine the pH and concentrations of H2A, HA, and Mapd A2- in a solution of 0.00195 M monopotassium fumarate (KHA). The pK, values for fumaric acid are 3.02 (pKat) and 4.48 (pKa2). Number pH7.17 Number H,A|-| 1.38 × 10-7 | M Incorrect You have assumed that the formal concentration of the solution 0.00195 M) is a good estimate of the concentration of the Number HA 0.00195 termediate species. When the formal concentration is small and the Ka values are nearly equal this assumption cannot be made. Repeat your calculations using a new estimate for the concentration of the intermediate species. This new estimate can be calculated with the following equation, Number [A1-]-195.48 A"-- 95.48 where F is the formal concentration, [H2A] is the concentration of the completely protonated species, [A2-1 is the concentration of the completely deprotonated species, and (HA is the concentration of the intermediate species. Continue to repeat the calculations in this manner until the pH is constant O Previous @ G ve t trt ,o

Explanation / Answer

First, assume the acid:

to be H2A, for simplicity, so it will ionize as follows:

H2A <-> H+ + HA-

HA- <-> H+ + A-2

where, H+ is the proton and HA- and A-2 are the conjugate bases, H2A is molecular acid

Ka1 = [H+][HA-]/[H2A]; by definition

Ka2 = [H+][A-2]/[HA-]

initially

[H+] = 0

[HA-] = 0

[H2A] = M;

the change

initially

[H+] = + x

[HA-] = + x

[HA2] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka1 = [H+][HA-]/[H2A]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.00195 M; then

x^2 + (10 ^-3.02)x - 0.00195 *(10 ^-3.02) = 0

solve for x

x =9.68*10^-4

substitute

[H+] = 0 + 9.68*10^-4 = 9.68*10^-4 M

[HA-] = 0 + 9.68*10^-4 = 9.68*10^-4 M

[H2A] = M - x = 0.00195-9.68*10^-4= 0.000982 M

Now...

Second ionization

HA- <-> H+ + A-2

initially

[H+] = x

[A-2] = 0

[HA-] = M-x

the change

[H+] = +y

[A-2] = + y

[HA-] = - y

in equilbrirum

[H+] = x+y

[A-2] = 0 + y

[HA-] = M - x - y

substitute in Ka

Ka2 = [H+][A-2]/[HA-]

Ka2 = (x+y)(y) / (M - x - y)

assume M-x >> M-x-y so (due to Ka2)

Ka2 = (x+y)(y) / (M - x)

y^2+ xy = Ka2(M-x)

y^2+ xy - Ka2(M-x) = 0

y^2+(9.68*10^-4 ) *y - (10^-4.48))*(0.000982) = 0

y = 3.25*10^-5

and we konw

[H+] = x+y =0.000982 +  3.25*10^-5 = 0.0010145

[A-2] = 0 + y =  3.25*10^-5

[HA-] = M - x - y = ( 0.000982- 3.25*10^-5) = 0.0009495 M

pH = -log(0.0010145) = 2.99

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