Question

(Answer from question 4)

HP− + H2O is in equilibrium with P2− + H3O+

This axid will attack the base

Assume 1 mol of KHP will neutralize 1 mol of NaOH

V = 25 ml

M = 0.1M

Mol Acid = mol Base

Mol base = M*V = 25 ml ^0.1 M = 2.5 mmol of base

therefore, you need 2.5 *10^-3 mol of acid, that is mol of KHP

Since KHP = 204.22 g/gmol

mass = mol*MW = 2.5*10^-3 * 204.22 = 0.51 grams needed

(Question 5)

With the above information explain exactly how you will perform this titration. Your explanation should include the calculations showing how much KHP should be weighed out as well as a description of the procedure including all glassware. Assume that approximately 25 mL of sodium hydroxide solution should be used

Answer 1

Solution :-

Since we know that moles ratio of the KHP and NaOH is 1 :1

And we have to use about 25 ml of 0.1 M NaOH solution.

Therefore moles of the amount of the KHP needed for the titration is calculated using the mole ratio of the NaOH and KHP

Moles of NaOH = M*V = 0.1 M * 0.025 L = 0.0025 mol NaOH

Therefore moles of KHP needed = 0.0025 mol because mole ratio is 1 : 1

Now lets convert moles of KHP to its mass

Mass = moles * molar mass

Mass of KHP = 0.0025 mol * 204.22 = 0.51 grams

Now lets write the procedure for the titration.

Equipment’s needed = burrete , pipette , volumetric flask, erlynmayer flask, phenolphthalein

First need to weight 0.51 g KHP and transfer to the 25 ml volumetric flask and then dilute it to the 25 ml mark of the flask.

Then fill the burette with the 25 ml NaOH solution and titrate it slowly with KHP solution in erlynmayer flask by adding the phenolphthalein indicator.

End point of the titration will be colorless to faint pink.