posted by Keith .
a .4000 M solution of nitric acid is used to titrate 50.00 mL of .237 M barium hydroxide (Assume that volumes are additive).
a) Write a balanced net ionic equation for the reaction that takes place during titration.
b) what are the species present at the equivalence point?
c) what volume nitric acid is required to reach the equivalence point?
d) what is the pH of the solution before any nitric acid is added?
e) what is the pH of the solution halfway to the equivalence point?
f) what is the pH of the solution at the equivalence point?
I said the net ionic equation was:
OH- + H+ --> H2O
And the species present are: Ba+2 and NO3-
I thought because, at the equivalence point only spectator ions are present, the equivalence point is 7. I was considering, for c), to say 7= pKa + log (base/acid), but since HNO3 is a strong acid, it wont have a Ka correct? So how do I approach this and letter d)?
a) is ok.
b) what you have answered is good; there will be a small amount of H^+ and OH^- there from the ionization of water but it is almost miniscule. (H^+)=(OH^-)= 1E-7 M. I don't know if that is neglibible in terms of your problem or not since I don't know the history of the class.
c)volume of HNO3 required to reach the equivalence point.
The equation is
2HNO3 + Ba(OH)2 ==> 2H2O + Ba(NO3)2
mols Ba(OH)2 = 0.05 L x 0.237 M = 0.01185.
mols HNO3 required = 0.01185 x (2 mols HNO3/1 mol Ba(OH)2) = 0.01185 x 2 = 0.02370.
L x M = mols; therefore, L = mols/M = 0.02370 mols HNO3/0.4000 = ?? Liters.
d)pH of Ba(OH)2 :
Ba(OH)2(aq) ==> Ba^+2 + 2OH^-
M Ba(OH)2 = 0.237 M = 0.237 mols/L. Therefore, (OH^-) = 2 x 0.237 = ??
pOH = - log (OH^-)=??
pH + pOH = pKw = 14
Solve for pH.
e). Halfway to the equivalence point:
You know the molarity of the Ba(OH)2 at the beginning. Twice that will be the (OH^-) at the beginning and that answer is the same as from part d. Then half that will be the (OH^-) when it is half gone. Calculate pOH and pH from that.
f). 7 is correct.