Of the following solutions, which has the greatest buffering capacity? 1.15 M HF
ID: 1014828 • Letter: O
Question
Of the following solutions, which has the greatest buffering capacity? 1.15 M HF and 0.624 M NaF 0.574 M HF and 0.312 M NaF 0.287 M HF and 0.156 M NaF 0.189 M HF and 0.103 M NaF They are all buffer solutions and would all have the same capacity. Calculate the percent ionization of nitrous add in a solution that is 0.260 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 times 10^4. 1.17 times 10^4 0.0450 4.16 0.314 5.78 What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The K_sp of PbCl2 is 1.6 times 10^5. 2.0 times 10^3 1.1 times 10^4 1.8 times 10^4 7.1 times 10^4 1.6 times 10^5 A 25.0 mL sample of a solution of a HCI is titrated with a 0.115 M NaOH. solution. The titration curve above was obtained. Which of the following indicators would be best this titration? methyl red (3.46) bromthymol blue (7.12) thymol blue (8.90) phenolphthalein (9.10) bromocresol purple (6.12)Explanation / Answer
36. The answer is A (1.15 M HF and 0.62M of NaF )
Whatever the concentration of HF. the buffer capacity is only determined by the concentration of added acid or base. more the concentration of acid or base more its buffer capacity
NaF is the strong base
Definition
Buffering capacity refers to the amount of added acid or added base that can be neutralized by a
buffer. It is determined by the concentrations of the conjugate acid and conjugate base. Buffering
capacity increases as these concentrations increase.
37. answer is d) 0.314
HNO2 <----> H+ + NO2-
HNO2
H+
NO2-
Initial
0.26
0
0
Change
-X
X
X
equllibrium
0.26-X
X
X
Dissociation consant = [product]/[reactant]
K = [NO2- ][ H+] /[ HNO2]
4.5 x 10-4= X*X /0.26-X
= X2/0.26- X
0.26 x 4.5 x 10-4 + 4.5 x 10-4 X =X2
X2 -4.5 x 10-4 x - 1.1205X10-4 = 0
Use quadratic equation to solve
x = 1.3x10-4 M
0.26-X = 0.26- 0.00013
=0.2599
0.2599 x 100 / 0.26 = 99.96%
38. answer is d) 7.1x10-4
Ksp = [Pb2+] [cl-]2
=
1.6X10-5 = [S] [2S]2
S is the solubility of lead chloride.
However, here the concentration of chloride anion has been (artificially) increased by the presence of hydrochloric acid, which gives stoichiometric quantities ofCl. So
[Cl] =0.15mol/lit
Ksp=[S][0.15+2S]2. Now we can (reasonably) make the approximation,[0.15+2S]0.15; we have to justify this approximation later.
So now,
Ksp[S][0.15]2, andS=Ksp(0.15)2=1.6×105/(0.15)2 = 7.1X 10-4
39. Phenolphthalein
Every strong acid Vs Storng base titration either methyl orange or Phenolphthalein was used as a indicator.
HNO2
H+
NO2-
Initial
0.26
0
0
Change
-X
X
X
equllibrium
0.26-X
X
X
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