Titrations problem?

Question

a) You will dissolve 0.30 g of potassium hydrogen phthalate (KHP, molar mass 204.22 g/mol) in 20 mL of water and titrate with the NaOH. To the nearest mL, calculate the volume of NaOH expected assuming that the NaOH concentration is ~0.1 M.

b) To 2 significant figures, what is the concentration of phthalate at the equivalence point? Recall that the total volume of solution is equal to the 20 mL used to dissolve the KHP plus the volume of the NaOH calculated in part a.

c) Given that Ka for hydrogen phthalate is 3.9 x 10-6 (or pKa = 5.41), what is Kb for phthalate to 2 significant figures? (hint: Ka*Kb = Kw = 10-14)

d) What is the pH at the equivalence point? (2 significant figures)


e) Is the pH in part d within the color change scheme for phenolphthalein?
a. Can't tell
b. No
c. Yes

Answer 1

a) in this case KHP acts as a monoprotic acid.
Thus the stoichiometry is 1:1 which means
mole KHP =mole NaOH =>
mass (KHP)/ MW (KHP)= M(NaOH)*VNaOH =>

V= mass/ (MW*M) = 0.3/(204.22*0.1) = 0.01469 L = 14.69 mL =15 mL

b)Vtotal =20+15= 35 mL =0.035 L. At the equivalence the total phtalate is equal to the KHP that you had (all of it reacts at the equivalence point)
C= mole/V= mass/(MW*Vtotal)= 0.3/(204.22*0.035) = 0.042 M

HOWEVER since pthalate (P-(2)) comes from a weak acid it will hydrolyze according to the reaction
P(-2) + H2O <=> HP(-) +OH(-)
So in order to find the true concentration of P(-2) at equilibrium at the equivalence point, you need to know the Kb. SO in fact you need to solve first (c), unless they were asking you for the total/initial pthalate concentration

Kb=Kw/Ka =(10^-14)/(3.9*10^-6) =2.6*10^-9

so
.. .. .. .. .. .. P(-2) + H2O <=> HP(-) +OH(-)
Initial .. .. .. ..C
React .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .x .. .. .. x
At Equil. .. C-x .. .. .. .. . .. .. .. .. x .. .. .. x

Kb= [OH-][HP-]/[P-2] = x^2/(C-x)
but we calculated that C=0.042 and Kb =2.6*10^-9 so
x^2/(0.042-x) = 2.6*10^-9

Let's assume that 0.042 >>x so that 0.042-x=0.042. Then the equation is simplified to x^2/0.042= 2.6*10^-9 =>
x=squareroot (0.042*2.6*10^-9) = 1.044 *10^-5 =1.0*10^-5 (2 significant figures). It is truly <<0.042 thus our assumption is valid (otherwise you would solve the quadratic for an exact solution)

because x is very small and they ask you to use 2 significant figures, the concentration of pthalate at equilibrium is practically 0.042 M (it is the assumption we made, but also 0.042-0.00001=0.04199 which is written as 0.042 using 2 significant figures)

pH=14-pOH= 14-(-logx)= 14-(-log(10^-5)) =14-5 =9.0 (2 significant figures).

Phenolpthalein has the pink colour at pH=8.2-10.0
below pH=8.2 the solution is colourless. KHP is acidic so you start with a colourless solution and you reach the equivalence point at pH=9.0. So yes, the pH at the equivalence point is within the color range scheme.