Untitled - Kelly Walsh High School

Buffers and Other Equilibria 247
pKb log Kb 10.65
pKb 1010.65 2.2387 1011 10 (unrounded)
pKb The K b equilibrium reaction is:
NO 2 (aq) H2O(l) K OH (aq) HNO 2(aq)
This leads to the K b relationship:
[OH ] [HNO2 ]
[NO2 ]
The ICE table for this equilibrium is:
K b 2.2387 10 11
NO 2 (aq) OH (aq) HNO2(aq)
Initial 0.100 0 0
Change x x x
Equilibrium 0.100 x x x
We can now enter the equilibrium line of the table into the K b expression and
solve for x:
[OH ] [HNO 2 ]
Kb 2.2387 1011 [x] [x]
[NO2 ] [0.100 x]
x 1.4962 10 6 M OH (unrounded)
Either we can use the hydroxide ion with the K w to find the hydrogen ion concentration,
and then find the pH, or we can find the pOH and use the pK w to
find the pH. We will use the latter method:
pOH log [OH ] log (1.4962 10 6 ) 5.8250 (unrounded)
pK w pH pOH 14.00
pH pK w pOH 14.00 5.8250 8.1750 8.18
d. 55.00 mL This must be in the fourth region of the titration curve. (We know
this is true because the preceding part gave us the equivalence point.) After the
equivalence point, the substance added will be excess and the other substance
is limiting. In this case, the sodium hydroxide will be in excess and the nitrous
acid will be limiting.