Question

For the next problem consider .40M H2A(aq) with Ka=1.0*10^-7 Ka2=5.0*10^-12

35) Ka,1/Ka,2 > 10^3 the weak diprotic acid approx

a. is not valid and we can use just te first step

b.is not valid and we must use both steps

c.is valid and we can use just the first step

d.is valid and we must use both steps

In the intial row of the equil. table (H^1+) is only approx. 0

a. because all intial vaues are approximate

b.because of the authroization of water

c. because of the leveling effect

d. because of charge dispersal

what is the pH of the solution?

a. 3.70

b.-2.70

c.7.00

d.4.30

what is [a^2-]

a. 1.0x10^-7

b.5.0*10^-5

c..40

d. 5.0*10^-12

Answer 1

Q1.

KA1/KA2 = (10^-7)/(5*10^-12) = 20000

since 20000 > 10^3

then, we can ignore the second dissociation... meaning the thumb rule is valid and we can use just first ionization

c.is valid and we can use just the first step

Q2

b.because of the authroization of water

since H2O actually will ionize as:

H2O + H2O <-> H3O + + OH-

so

[H+] = 10^-7 approx, which is neglegible

The pH of the solutoin, before any addition:

pH = 7 since it is pure water

after acid addition

KA1 = [H+][HA-]/[HA-]
10^-7 = (x*x)/(0.4-x)

x = 1.99*10^-4

pH = -log(1.99*10^-4) = 3.701 in equilbirium; choose a

finally

[A-2]:

KA2 = [H+][A-2]/[HA-]

5*10^-12 = (1.99*10^-4+x) (x) / (1.99*10^-4-x)

solve for x

(5*10^-12) (1.99*10^-4) = (1.99*10^-4)x + x^2

x^2 + 1.99*10^-4x - 9.95*10^-16 = 0

x = 4.99*10^-12

[A] =  4.99*10^-12 M