A student is asked to calculate the volume of 0.10M NaOH that must be added to 60 mL of 0.20M acetic acid (Ka=1.8e-5) in order to produce a solution for which [H+]=9.0e-6M. Which of the following volumes is closest to that which she will need?
b. 60 mL
The answer isnt given either and there is no explanation so I am not sure where to start.
acid dissociation: HA + H2O <-> A- + H+
acid dissociation equilibrium constant: Ka=(H+ * A-)/HA
plug in givens: (1.8e-5)=(9e-6)*(A-/HA)
divide both sides by (9e-6): 2=A-/HA
You start with 0.06*0.2=0.012 moles of HA, and no A-. As you add strong base you lose HA by (0.012-x) and gain A- by (x)
now have: 2=(x)/(0.012-x)
solve for x [I did trial and error to see (8)/(12-8)]= 8/4=2): x=0.008
x is the moles of NaOH added so solve for volume: 0.1*V=0.008 -> V=80 mL