posted by Tiffany on .
Suppose that 55 mL of 0.2 M LiOH and 55 mL of 0.3 M HNO2 are mixed.
What is the K for the dominant equilibrium in terms of Ka's, Kb's, Kw etc.?
LiOH + HNO2 ==> LiNO2 + H2O
mmoles LiOH = 55 x 0.2 = 11 mmoles.
mmoles HNO2 = 55 x 0.3 = 16.5 mmoles.
The HNO2 is in excess so you will have remaining 11 mmoles LiNO2, 11 mmols H2O, and 4.5 mmoles HNO2 left. So the HNO2 will ionize and the dominant equilibrium will be HNO2 + H2O --> H3O^+ NO2^- and
Ka = (H3O^+)(NO2^-)/(HNO2)