Practice Problem Set: Part A: Standardization of a Base USE THE FOLLOWING DATA T

Question

Practice Problem Set: Part A: Standardization of a Base USE THE FOLLOWING DATA TO CALCULATE THE MOLARITY OF THE SODIUM HYDROXIDE SOLUTION. SHOW YOUR CALCULATIONS CLEARLY. THE MOLAR MASS OF KHP = 204.22 g/mol The following data represent data from the two best runs. Mass KHP (g) 0.7621g 0.7549g Initial Buret Reading (mL) 0.00 mL 14.95 mL Final Buret Reading (mL) 14.95 mL 29.70 mL Volume of NaOH used ______ ______ [NaOH] _______ _______ You should calculate the % difference between the 2 results Part B: Determination of the mola rity of a Polyprotic Acid. The standard ized base form Part A is used to determine the molari t y of an polyprotic acid. A 15.00 mL sample of acid is titrated with the standard base . If the buret readings are V i = 0.25 mL and V f = 30.15 mL, what is the concentration of the unknown acid if its structure is? 1. HA [HA] = _______ M 2. H 2 A [H 2 A] = _______ M 3. H 3 A [H 3 A] = ________ M

Explanation / Answer

Part-A:

The balanced equation for the reaction of NaOH with KHP, KHC8H4O4 is

NaOH(aq) + KHC8H4O4(aq) ----------- > KNaC8H4O4(aq) + H2O(l)

1mol, --------- 1 mol, --------------------------- 1 mol

From the above balanced chemical reaction it is clear that 1 mol NaOH reacts with 1 mol KHP.

1st run:

mass of KHP = 0.7621 g

=> moles of KHP = mass/molar mass = 0.7621 g /  204.22 g/mol = 3.732x10-3 mol

Volume of NaOH used, V = 14.95 mL - 0.00 mL = 14.95 mL = 0.01495 L

Let the concentration of NaOH be 'C1' mol/L

moles of NaOH = moles of KHP

=> C1 x V(L) = C1 x 0.01495L = 3.732x10-3 mol

=> C1 = 3.732x10-3 mol / 0.01495L = 0.2496 M

2nd Run:

mass of KHP = 0.7549g

=> moles of KHP = mass/molar mass = 0.7549 g / 204.22 g/mol = 3.6965x10-3 mol

Volume of NaOH used, V =29.70 mL -14.95 mL= 14.75 mL = 0.01475 L

Let the concentration of NaOH be 'C2' mol/L

moles of NaOH = moles of KHP

=> C2 x V(L) = C2 x 0.01475L = 3.6965x10-3 mol

=> C2 = 3.6965x10-3 mol / 0.01475L = 0.2506 M

Average of the concentration of C1 and C2 = (0.2506 + 0.2496) / 2 = 0.2501 M

Percentage difference between the two results =[(C2-C1) / C2] x 100

=[(0.2506 - 0.2496) / 0.2506] x 100 = 0.400 % (answer)

Part-B:

1: For monoprotic acid, HA:

Volume of NaOH used = 30.15 mL - 0.25 mL = 29.9 mL = 0.0299 L

Concentration of standard NaOH = 0.2501 M (calculated from part-A)

Hence moles of NaOH used = MxV = 0.2501 mol/L x 0.0299 L = 0.007478 mol

Volume of HA = 15.00 mL = 0.015 L

For monoprotic acid, moles of NaOH = moles of HA

=> 0.007478 mol = 0.015 x C

=> C = 0.499 M

=> [HA] = 0.499 M (answer)

2: For Diprotic acid(H2A):

When we use a diprotic acid, 1 mol of H2A reacts with 2 mol of NaOH

=> [H2A] = 0.499M / 2 = 0.2495 M (answer)

Similarly when a triprotic acid is used, 1 mol of H3A reacts with 3 mol NaOH

=> [H3A] = 0.499M / 3 = 0.1663 M (answer)

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