What is the approximate ph at the equivalence point of a weak acid-strong base titration if 50.00 ml of aqueous hydrofluoric acid requires 60.00 ml of 0.200 mlioh? ka = 6.76×10−4 for hf.

 

HF + NaOH = F- + H2O 

At the equivalence point all the acid (HF) is converted into its conjugate base. 

So basically we need to calculate the amount of the conjugate base which equals the number of moles of OH-. 

We know at equivalence point moles of OH- (from NAOH) = moles of F- 

moles of OH = .030 L * .4 M = .0012 moles OH- 

Thus .0012 moles of OH consumed the weak acid HF and produced .0012 moles of F- 

Now we need to calculate the concentration of F- = .0012/(.030 + .025) 

2.18E-2 M. 

Now let's write another equation to indicate the equilibrum established between F- and HF. 

F- + H2O <> HF + OH- 

The dissociation constant (Kb) = [HF][OH-]/[F-] 

We ignore water. 

If X amount of F- dissociates than X amount of HF and OH are produced. 

Kb = 1/Ka = [X][X]/[2.18E-2 - X] 

Since 2.18E-2 >>> X we can ignore X in the denominator 

Kb = 1/Ka = [X]^2/[2.18E-2]. 

Solve for X will give you the [OH-] concentration 

Solve for pOH (pOH = -log[OH-] 

pH = 14-pOH.

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