Question

Acid / Base Titrations: Pre-Laboratory Assignment 1. Give at least two characteristics desirable in a primary standard. 2. Define equivalence point and ritration and point. Do these two points in a titration occur when the same volume of titrant has been added? Explain 3. A student standardized a NaOH solution using KHC.H.O.. Three different KHC&H.O. samples were dissolved in water and titrated with the same NaOH solution, each to a phenolphthalein end point. The pertinent reaction is: KHC H404 aq) + NaOH/ aq) + NaKC4H404 aq) + H20(4) Results from these three runs are summarized in the data table below. Complete the following summary table of calculated values using the given experimental measurements. Mass of KHC.H.O. titrated, 0.452 0.470 0.442 Volume of NaOH(aq) required, mL 15.45 15.99 15.05 Number of moles of KHC.H.O.titrated, mol Number of moles of NaOH required, mol Volume of NaOH solution required, L Molarity of NaOH solution, mol/L Mean molarity of NaOH solution, mol/L Show clear work for the following: a) Moles of KHC.H.Otitrated for Run #1 b) Moles of NaOH required for Run #1 c) Molarity of NaOH as determined by Run #1 Acid Base Titration p. 5

Answer 1

1. Desirable characteristics of primary standard are high purity, low reactivity, high stability, low hygroscopicity (should not absorb moisture from surrounding), non-toxic and should be readily available.

2. Equivalence point is different from end point of the titration.

When the number of moles of both acid and base are present in equal concentration, it is known as equivalence point.

When the indicator changes its colour, it is known as end point. Volume of base in the solution is more at end point (usually 1-2 drops more).

3. Calculations for Run #1:

Mass of KHC₈H₄O₄ = 0.452 g

Molar mass of KHC₈H₄O₄ = 204.22 g/mol

So, # moles of KHC₈H₄O₄ = given mass ÷ molar mass

# of moles = 0.452 ÷ 204.22 = 0.0187 mol

Now, according to the balanced chemical equation, 1 mol KHC₈H₄O₄ reacts with 1 mol of NaOH.

Therefore, 0.0187 mol of KHC₈H₄O₄ must react with 0.0187 mol of NaOH.

Volume of NaOH (in L) = 15.45/1000 = 0.01545 L

Now, Molarity = # of moles ÷ Volume (in L)

=> Molarity = 0.0187 mol ÷ 0.01545 L = 1.210 M.