Question

If 0.010 mol of sodium hydroxide is added to 1L of a buffer that is 0.200 M HClO2 and 0.155 M NaClO2, what is the pH? Ka for HClO2 is 1.1 x 10^-2

A. 2.07
B. 1.85
C. 1.96
D. 1.90
E. 2.01

Answer 1

millimoles HClO2 = 1000 mL x 0.200 M = 200
millimoles NaClO2 = 1000 mL x 0.155 M = 155
moles NaOH added = 0.010 = 10 millimoles.
...............HClO2 + NaOH ---> NaClO2 + H2O
I...............200...........0...................155..............
add...........................10.......................................
C.............-10.............-10................+10....................
E...............190..............0..................165......................
Convert Ka for HClO2 to pKa. pKa = -log Ka then
pH = pKa + log [(base)/(acid)]
pH = pKa + log [(NaClO2)/(HClO2)]
Plug in from the E line and calculate. I want you to realize that technically you should use concentrations; however, you may use millimoles and get the same answer faster. Why? Because (NaClO2) = millimoles/mL = 165/1000 and
(HClO2) = 190/1000 and volume ALWAYS will cancel.
Post your work if you get stuck.