In this experiment, i have to titrate a known concentration of the weak acid pot

Question

In this experiment, i have to titrate a known concentration of the weak acid potassium hydrogen phthalate (KHP) with a strong base, NaOH. From the titration curve, which is a plot of the pH veruss the volume of the NaOH added (which is the X axis), I need to determin the acid ionization constant (Ka) for KHP.

The weak acid KHP is a salt that dissociates in water

KHC8H4O4 (s)--> K+ (aq) + HC8H4O4- (aq)

It is an anion (HC8H4O4) that behaves like an acid:

HC8H4O4- + H2o <--> H3o+ + C8H4O4        Ka for KHP= [H3o][C8H4O4]/[HC8H4O4]

When titrated with NaOH (strong base), the NaOH will react with the acid:

NaOH + HC8H4O4 --> H20 + NaC8H4O4

We have 0.1 M of NaOH, and we use 20 ml of NaOH to get to the equivalence point. We use 0.4 g KHP.

a). from the titration cuve (NaOH added on x axis, recorded pH on the Y axis), determine the equivalence point and helfway point

b). from the halfway point, determin the Ka for KHP

c). from the equivalence point, determine the exact concentration of NaOH

d). Using the Ka and the known concentration of KHP, calculate the initial pH of the acid and compare to the experimental value

Explanation / Answer

Part a)

pH at the equivalence point = 8.6

Part b)

At the halfway,

pH = pKa + log(salt/acid)

Here, [salt] = [acid]

Therefore,

pH = pKa = 5.4

Part c)

We have, molar mass of KHP = 204.22 g/mol

Initial moles of KHP = 0.4/204.22 = 1.96 x 10-3

At the equivalence point, number of moles of KHP = Number of moles of NaOH

Therefore, concentration of NaOH = (1.96 x 10-3) /(20 x 10-3) = 0.098 M

Part d)

Concentration of KHP be x.so,

pH for the salt of strong base and a weak acid can be calculated using,

pH = 7 + 0.5 pKa + 0.5log(C)

Here, C is the concentration of the salt

8.6 = 7 + (0.5 x 5.4) + 0.5 x log(0.098 x[20/(20+x)])

Ons olving,

x = 297 x 10-3 L

Therefore, concentration of KHP = (1.96 x 10-3)/(297 x 10-3) = 6.59 x 10-3 M

Let us assume x be the dissociation of KHP,

x = [H+]

Therefore,

10-5.4 = x2/(6.59 x 10-3 M-x)

On solving,

x = [H+] = 1.6 x 10-4 M

pH = -log(1.6 x 10-4 M) = 3.79

Experimental value is very close, which is 4.1

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