Use the table below to answer questions 2 through 5. HCN HF HNO2 HIO 4.9 x 10-0
ID: 555273 • Letter: U
Question
Use the table below to answer questions 2 through 5. HCN HF HNO2 HIO 4.9 x 10-0 7.1 x 10 4.5 x 10 2.3 x 10 2) Which acid in the table above will have the strongest conjugate base? Hint, look for the conjugate base that will have the biggest Kb. 3) A 0.500 M solution of potassium cyanide (KCN) is prepared. Determine the pH of this solution. 4) Which acid in the table above could be used with its conjugate base to prepare a buffer with a pH of 9.3? 5) Determine the pH of a buffer made with 1.00 molar HF and 0.500 M NaF.Explanation / Answer
Q2.
strongest conjguate base --> will be the weakest acid, since the base will be stronger (counterbalance)
from the list
HIO = H+ + IO- is the strongest base, which will form HIO readily
Q3
the pH of KCN
expect hydrolysis
Let HA --> HCN and A- = CN- for simplicity
since A- is formed
the next equilibrium is formed, the conjugate acid and water
A- + H2O <-> HA + OH-
The equilibrium s best described by Kb, the base constant
Kb by definition since it is an base:
Kb = [HA ][OH-]/[A-]
Ka can be calculated as follows:
Kb = Kw/Ka = (10^-14)/(4.9*10^-10) = 2.04*10^-5
get ICE table:
Initially
[OH-] = 0
[HA] = 0
[A-] = M
the Change
[OH-] = + x
[HA] = + x
[A-] = - x
in Equilibrium
[OH-] = 0 + x
[HA] = 0 + x
[A-] = M - x
substitute in Kb expression
Kb = [HA ][OH-]/[A-]
2.04*10^-5 = x*x/(0.5-x)
solve for x
x^2 + Kb*x - M*Kb = 0
solve for x with quadratic equation
x = OH- =0.00318
[OH-] =0.00318
pOH = -log(OH-) = -log(0.00318= 2.50
pH = 14-2.50 = 11.5
pH = 11.5
Q4.
for a buffer
pH = pKa + loG(A-/HA) is recommended
also, A-/HA must be nearest to 1, so
pH = pKa goal
Ka = 10^-pH= 10^-9.3 = 5.01*10^-10
from the list, the most recommended is HCN
Q5
ph of a buffer given
pH = pKa + log(NaF/HF)
pKa = -log(7.1*10^-4) = 3.15
pH = 3.15 + log(0.5/1)
pH = 2.8489
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