The indicator methyl red has a pKhln=4.95 . It changes from red to yellow over the pH range from 4.4 to 6.2.
If the indicator is placed in a buffer solution of Ph = 4.56, what percent of the indicator will be present in the acid form,Hln- , AND what percent will be present in the base or anion form, ln-?
If someone could please help me out, thanks in advance!!
I would use the Henderson-Hasselbalch equation.
4.56 = 4.95 + log (base)/(acid)
Solve for (B/A) ratio. Then
A + B = 100 (100 is just a convenient number)
Two equations in two unknowns. Solve for A and B and these will be in percent since we chose 100 as the total number of A+B.
Sure, I'd be happy to help! But before I give you an answer, I have a quick question for you - why did the scarecrow win an award?
To determine the percent of the indicator in the acid form, Hln-, and in the base or anion form, ln-, at a pH of 4.56, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([ln-] / [Hln-])
First, let's determine the values of [ln-] and [Hln-].
Given:
pH = 4.56
pKa = 4.95
From the equation above, we can rearrange it as follows:
[Hln-] / [ln-] = 10^(pH - pKa)
[Hln-] / [ln-] = 10^(4.56 - 4.95)
[Hln-] / [ln-] = 10^-0.39
[Hln-] / [ln-] ≈ 0.412
Now, let's assume we have x moles of the indicator in total. Then, we can express [ln-] and [Hln-] in terms of x:
[ln-] = 0.412x
[Hln-] = (1 - 0.412)x = 0.588x
The percent of the indicator in each form can be calculated as follows:
Percent Hln- = ([Hln-] / x) * 100 = (0.588x / x) * 100 = 58.8%
Percent ln- = ([ln-] / x) * 100 = (0.412x / x) * 100 = 41.2%
Therefore, at a pH of 4.56:
- 58.8% of the indicator will be present in the acid form, Hln-
- 41.2% of the indicator will be present in the base or anion form, ln-
To determine the percent of the indicator present in the acid form (Hln-) and the base form (ln-), you can use the Henderson-Hasselbalch equation. The equation relates the pH of the solution to the ratio of the concentrations of the acid (Hln-) and its conjugate base (ln-).
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([ln-]/[Hln-])
Where:
- pH is the pH of the solution
- pKa is the negative logarithm of the acid dissociation constant of the indicator (pKhln, in this case)
- [ln-] is the concentration of the base form
- [Hln-] is the concentration of the acid form
First, we need to determine the ratio [ln-]/[Hln-] using the Henderson-Hasselbalch equation. Rearranging the equation gives:
[ln-]/[Hln-] = 10^(pH - pKa)
Substituting the given values:
- pH = 4.56 (from the buffer solution)
- pKa = 4.95 (given for the indicator)
[ln-]/[Hln-] = 10^(4.56 - 4.95) = 0.403
The ratio [ln-]/[Hln-] is 0.403. This means that for every 1 part of the indicator as the acid form (Hln-), there will be approximately 0.403 parts of the indicator as the base form (ln-).
To determine the percentages, we can convert the ratio to percentages:
Percent of Hln- = [Hln-] / ([Hln-] + [ln-]) * 100
Percent of ln- = [ln-] / ([Hln-] + [ln-]) * 100
Since [ln-] / [Hln-] = 0.403, we can assign any value to [Hln-] (e.g., 1) and find the corresponding value of [ln-].
Percent of Hln- = 1 / (1 + 0.403) * 100 ≈ 71.32%
Percent of ln- = 0.403 / (1 + 0.403) * 100 ≈ 28.68%
Therefore, approximately 71.32% of the indicator will be present in the acid form (Hln-), and approximately 28.68% will be present in the base form (ln-).