23) Answer: 0.06660 grams b) What was the pH of the solution at the beginning of
ID: 953017 • Letter: 2
Question
23)
Answer: 0.06660 grams
b) What was the pH of the solution at the beginning of the titration (assuming no contribution from the inert material).
Answer: pH = 5.27
c) What was the pH of the solution after the addition of 5.00 mL of 0.1000 M NaOH (assuming no contribution from the inert material)?
Answer: pH = 9.07
d) What was the pH of the solution at the equivalence point (assuming no contribution from the inert material)?
Answer: pH = 10.88
I just want to see how they got all of that.
23. A sample ofammonium chloride (NH4Cl, MW. 53.49 g/mol; Ka, NH,'-570 x 10"o) was contaminated with inert material. The whole sample was dissolved in 25.00 mL of water, and this solution was titrated with 12.45 mL of 0.1000 M NaOH to reach the end point. a. (10 points) Calculate the amount of ammonium chloride present in the sample (in grams)Explanation / Answer
For letter a)
We know we have, due to NaOH added:
0.01245 L * 0.1 mol/L = 0.001245 moles of NH4Cl
Converting to grams:
0.001245 moles of NH4Cl * (53.49 g/mol) = 0.0666 grams
For letter b:
We get NH4+ ion concentration:
Ka = [NH4+][Cl-] / [NH4Cl]
5.7 x 10-10 = x2 / (0.001245/0.025)
x = 5.32785 x 10-6 M = [H+]
pH = -log(5.32785 x 10-6) = 5.27
For letter c)
NH4Cl + NaOH -> NH4OH + NaCl
pH = pKa + log [Base/Acid]
We determine quantity of NH4Cl:
0.01245 L * 0.1 mol/L = 0.001245 moles of NH4Cl
If we add only 5 mL
0.005 L * 0.1 mol/L = 0.0005 moles of NaOH
Moles remaining = 0.001245 - 0.0005 = 0.000745 moles
New concentrations:
[Acid] = 0.000745 moles / 0.03 L = 0.02483 M
[Base] = 0.0005 moles / 0.03 L = 0.01667 M
And to get pH:
pH = -log(5.70 x 10-10) + log [0.01667/0.02483]
pH = 9.07
For letter d)
We get concentration of NH4OH, as it is the product we get at equivalence:
0.001245 moles of NH4OH produced / 0.03745 L = 0.033 M
As it is a weak base, we get Kb:
Kb = Kw/Ka
Kb = 1x10-14 / 5.7x10-10 = 0.0000175
As defined for constant Kb in dissociation:
Kb = [NH4+][OH-] / [NH4OH]
0.0000175 = x2 / 0.033 M
x = 0.0007599 = [OH-]
pOH = -log (0.00076) = 3.12
pH = 14 - 3.12 = 10.88
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