Question

A. At pH 7, what are the relative concentrations of ionized and un-ionized p-nitrophenol? (pKa of p-nitrophenol is 7.0)

B. If enough concentrated hydrochloric acid is added to a solution of p-nitrophenol to lower the pH from 7 to 5, what will happen to the relative concentrations of the ionized and un-ionized forms?

C. Ionized p-nitrophenol has a yellow color, while the un-ionized form is colorless. The yellow color can be measured using a spectrophotometer at 400nm. In order to determine the total amount of p-nitrophenol in a solution, would you perform the spectrophotometer reading at an acidic or basic pH? Clearly explain why?

D. A solution of p-nitrophenol at pH 8.65 was found to have an A₄₀₀ of 0.470. What is the total concentration (in µM) of p-nitrophenol (ionized plus un-ionized) in the solution? The molar extinction coefficient of p-nitrophenol is 17,500 M^-1cm^-1 and the pKa is 7.

Answer 1

A. At pH 7, what are the relative concentrations of ionized and un-ionized p-nitrophenol? (pKa of p-nitrophenol is 7.0)

if pKa = 7, then expect, from the buffer eqiation

HA <-> H+ + A-

pH = pKa + log(a-/HA)

7 = 7 + log(A-/HA)

if

A- = HA, then 50% is ionized, 50% is not ionized

B. If enough concentrated hydrochloric acid is added to a solution of p-nitrophenol to lower the pH from 7 to 5, what will happen to the relative concentrations of the ionized and un-ionized forms?

note that if there is more H+ present

the next equilbirium is expected:

HA <-> H+ +A-

if H+ is added, expect much more HA (not ionized) than A- (ionized)

C. Ionized p-nitrophenol has a yellow color, while the un-ionized form is colorless. The yellow color can be measured using a spectrophotometer at 400nm. In order to determine the total amount of p-nitrophenol in a solution, would you perform the spectrophotometer reading at an acidic or basic pH? Clearly explain why?

ionized = yellow, unionized = colorless

in order to get a yello wcolor, there must be plenty of A- present ( which is yellow)

therefore, it must be basic, i.e. higher than pH >7

D. A solution of p-nitrophenol at pH 8.65 was found to have an A400 of 0.470. What is the total concentration (in µM) of p-nitrophenol (ionized plus un-ionized) in the solution? The molar extinction coefficient of p-nitrophenol is 17,500 M-1cm-1 and the pKa is 7.

Apply:

A = e*l*C

C = A/(e*l)

C = 0.47 / (17500*1) = 0.000026857 M

so

pH = pKa + log(A-/HA)

8.65 = 7 + log(0.000026857 / x)

x = 0.000026857 / (10^(8.65-7)) = 6.012*10^-7

total = 0.000026857 + 6.012*10^-7 = 0.00002745 M = 0.00002745*10^6 mM = 27.45micro M