Question

1) Explain why a primary standard, like KHP, must be used to standardize a sodiumhydroxide solution. Define hygroscopic.

2) Determine the molarity of NaOH if 22.05 mL of NaOH is used to titrate 0.3440 g of KHP.

3) What is the molarity of a monoprotic acid if 19.00 mL of 0.102 M NaOH is used to titrate 25.00 mL of the acid?

4) What is the Ka of an acid if the pH of the solution at the half-point is equal to 5.8?

5) Write the balanced reactions for this experiment. (see link --> https://drive.google.com/file/d/0B6dlfmHWuu5taVhrZTREeHZMUnM/view?usp=sharing )

Answer 1

1) Neccessity of Standardization of NaOH solution :

Potassium Hydrogen Pthalate (KHP) is a primary standard solution which means that onse the solution of KHP is prepared of known molarity it won't change over very long period of time.

But NaOH (s) is a Hygroscopic solid which very often have water adhered on. Hence while weighing the NaOH pelletes for preparation of NaOH solution there introduce an error in calculated mass taken for preparation of aq. NaOH solution of known molarity. This error in mass calculation changes the molarity of solution formed.

Thus an aq. NaOH solution formed has no fixed or accurate molarity.

To know the exact molarity of solution formed we titrate this solution with unknown molarity with the one with known molarity.

This is known as Standardization of solution.

Standardization gives the exact concentration of the solution at that instant.

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2) Let us first calculate moles of KHP in 0.344 g KHP.

Molar mass of KHP= 204.22 g/mole.

Moles of KHP in 0.344 g = 0.344 / 204.22 = 1.68 x 10^-3 moles.

Now, for NaOH, Let molaity be M and volume required is 22.05 mL = 22.05 x 10^-3 L

So, Moles of NaOH = Molarity x volume = M x 22.05x10^-3 moles.

KHP is a monoprotoc acid and NaOH is monoacidic base and both are strong electrolytes.

Hence, at neutralization,

Moles of NaOH = Moles of KHP

M x 22.05x10^-3 = 1.68 x 10^-3.

M x 22.05 = 1.68

M = 1.68 / 22.05

M = 0.076 M

Molarity of NaOH is 0.076 M

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3) For a titration of monoprotic acid, say Molarity = M and Volume (V1) = 25.00 mL

And for Monoacidc base NaOH, Molarity (M2) = 0.102 M Volume (V2) = 19.00 mL

At neutralization point,

Milimoles of Acid = Milimoles of Base

MV1 = M2V2

Mx25.00 = 0.102 x 19.00

M = 0.102 x 19.00 / 25.00

M = 0.078 M

Molarity of unknown acid is 0.078 M.

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4) At half point of neutralization,

pKa = pH

Given pH = 5.8

So, pKa = 5.8

By definition,

pKa = -logKa

Ka = 10^-pKa.

Ka = 10^-5.8.

Ka = 1.58 x 10^-6.

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