For the next problem consider .40M H2A(aq) with Ka=1.0*10^-7 Ka2=5.0*10^-12 35)
ID: 499515 • Letter: F
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
Explanation / Answer
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