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crossled

2+ Year Member
Aug 26, 2015
25
16
Status
Medical Student
Thanks for the reply. The problem is:

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?

a. 30mL
b. 60 mL
c. 80mL
d. 120mL


The answer isnt given either and there is no explanation so I am not sure where to start.
 
Last edited:

Neuroplasticity

2+ Year Member
Jul 27, 2015
102
53
Status
Medical Student (Accepted)
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