The indicator methyl orange has a pK_a = 3.46. It changes color from red to oran
ID: 1051330 • Letter: T
Question
The indicator methyl orange has a pK_a = 3.46. It changes color from red to orange over the pH range 3.1 to 4.4. If methyl red is placed in a 1.00 liter buffer solution that is 0.250 M HNO_2 and 0.150 M NaNO_2, what percent of the indicator will be in the acid form? K_a(HNO_2) = 7.1 times 10^-4 Calculate what the pH of the solution would have to be in order for the concentration of the basic form of the indicator to be 20 times greater than the concentration of the acidic form. Calculate the mass of NaOH that would have to be added to increase the pH of the solution to its value in part b.Explanation / Answer
Solution:
Using Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Here the variables are In- and HIn to indicate the base form the indicator and the acid form of the indicator respectively. Since we need to find out the percentage of the indicator, we really want the ratio of conjugate base to acid.
pH = pKa + log([In-] / [HIn])
Here, pKa is the pKa of the indicator in question. Re-arranging the above equation we have:
pH - pKa = log([In-] / [HIn])
10(pH - pKa) = [In-] / [HIn] --------------------(1)
Now we need to find the pH of 1.00 L buffer of 0.25M HNO2 and 0.15M NaNO2 as follows:
NaNO2 ----------------> Na+ + NO2-
0 0.15 0.15
HNO2 ------------------> H+ + NO2-
0.25 M 0 0 (initial)
0.25 –x x 0.15+x
Now,
Ka = [H+][ NO2-]/[HNO2]
7.1x10-4= x(0.15+x)/(0.25-x)
7.1x10-4 = 0.15x/0.25; [neglecting x since HNO2 is a weak acid]
x = 0.25x7.1x10-4/0.15 = 11.83x10-4
Therefore, pH = -log (11.83x10-4) = 4 – 1.07 = 2.93
Now applying equation (1) we have:
10(pH - pKa) = [In-] / [HIn]
102.93 – 3.46 = [In-] / [HIn]
10-0.53 = [In-] / [HIn]
0.295 = [In-] / [HIn]
It means that at an acidic pH (= 2.93), there is only 29.5% of the conjugate base of the indicator (In-) and 70.5 % of HIn.
(b) We have: pH - pKa = log([In-] / [HIn])
When, [In-] = 20 [HIn], using above equation we have:
pH - pKa = log(20[HIn] / [HIn])
pH = log(20[HIn] / [HIn]) + pKa
or, pH = log 20 + pKa = log20 + 3.46 = 1.3 + 3.46 = 4.76
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