2. We have stated in class that in a \"Type 6\" problem like a titration in whic
ID: 1030324 • Letter: 2
Question
2. We have stated in class that in a "Type 6" problem like a titration in which different acids and bases are mixed together, you should first run a neutralization reaction to completion. Let's explore this idea and see if it is indeed a reasonable approach to this type of problem. a) fa strong acid and strong base are mixed together, we can describe this as H+ reacting with OH Write the reaction for the acid/base reaction that occurs when H+ is mixed with OH-and determine the equilibrium constant for this reaction. Is it reasonable to simply assume this reaction proceeds to completion? b) Now let's consider a reaction between a weak acid and a strong base. Suppose that a solution of HF is mixed with a solution of NaOH. Write the reaction for the neutralization reaction that occurs, and determine the equilibrium constant for this reaction. Is it still reasonable to simply assume that this reaction proceeds to completion?Explanation / Answer
2a) The balanced chemical equation for the reaction is given as
H+ (aq) + OH- (aq) ---------> H2O (l)
The equilibrium constant for the reaction is written as
K’ = [H2O]/[H+][OH-]
[H2O] is taken as unity in the expression, since H2O is an undissociated liquid (doesn’t break up to produce H+ and OH- ions). Therefore, the equilibrium constant is given as
K = 1/[H+][OH-] = 1/Kw
where Kw = [H+][OH-] = 1.0*10-14 is the ionic product of water. Therefore, we have
K = 1/(1.0*10-14) = 1.0*1014
The equilibrium constant for the reaction is unusually high indicating that the reaction indeed goes to completion.
2b) HF is a weak acid and ionizes partly in solution in produce H+ and F-, which is itself a weak base. The ionization reaction is given as
HF (aq) ------> H+ (aq) + F- (aq)
Ka = [H+][F-]/[HF]
NaOH, a strong base reacts with HF as below.
NaOH (aq) + HF (aq) -------> NaF (aq) + H2O (l)
H2O is undissociated as above. However, NaF is the sodium salt of the weak base F- which reacts with water to form HF as below.
F- (aq) + H2O (l) --------> HF (aq) + OH- (aq)
The equilibrium constant for the reaction is the base ionization constant of F- and is written as
Kb = [HF][OH-]/[F-]
Due to the two reactions shown above, the reaction doesn’t go to completion.
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