A certain indicator, HA, has a Ka value of 1.3 × 10-6. The protonated form of the indicator is blue and the ionized form is red.
What is the pKa of the indicator?
What is the color of this indicator in a solution with pH = 8? Blue, red, or purple?
Well, it seems like HA is quite indecisive about its color! But don't worry, I'm here to help you figure it out.
To find the pKa of the indicator, we need to take the negative logarithm (base 10) of its Ka value.
pKa = -log10(Ka)
So, let's crunch those numbers:
pKa = -log10(1.3 × 10-6)
*pummels calculator*
Ah-ha! According to my calculations, the pKa of the indicator is approximately 5.89.
Now, let's move on to the color party!
Since the pH is 8, we can compare it to the pKa to see if the indicator will predominantly exist in its protonated (blue) or ionized (red) form.
If the pH is greater than the pKa, it means the solution is more basic and the indicator will exist mostly in its ionized (red) form. So, in this case, the color of the indicator in a solution with pH = 8 will be... drumroll, please... RED!
Looks like this indicator is all about that reddish charm at pH 8.
To find the pKa of an indicator, you can use the equation pKa = -log(Ka), where Ka is the acid dissociation constant.
In this case, the Ka value is given as 1.3 × 10^(-6).
Using the equation, we can calculate the pKa as follows:
pKa = -log(1.3 × 10^(-6))
pKa ≈ -(-5.89)
pKa ≈ 5.89
Therefore, the pKa of the indicator is approximately 5.89.
To determine the color of the indicator in a solution with pH = 8, we need to compare the pH to the pKa.
If the pH is greater than the pKa, the indicator will be ionized and will exhibit its ionized form's color. In this case, the ionized form is red.
Since pH = 8 is greater than the pKa of approximately 5.89, the indicator will be ionized and will appear red in a solution with pH = 8.
So, the color of the indicator in a solution with pH = 8 is red.
To determine the pKa of an indicator, you can use the equation: pKa = -log(Ka), where Ka is the acid dissociation constant.
In this case, the Ka value is given as 1.3 × 10^-6. So, to find the pKa, we can perform the following calculation:
pKa = -log(1.3 × 10^-6)
Using a calculator, we can find that pKa is approximately 5.89.
Now, to determine the color of the indicator in a solution with pH = 8, we can compare the pH of the solution to the pKa of the indicator.
If pH > pKa, the indicator will be in its ionized form and will appear red.
If pH < pKa, the indicator will be in its protonated form and will appear blue.
If pH = pKa, the indicator will be in equilibrium between the two forms and will appear purple.
In our case, pH = 8, which is greater than the pKa of 5.89, so the indicator will be in its ionized form and appear red in a solution with pH = 8.
pKa=-log(Ka)=-log( 1.3 × 10-6)
pH=pKa=log(A-/HA) solve for the ratio
purple means ratio is one or very close to it
blue means ratio is 1>
red means ratio is 1<
****But check my thinking.