Methyl violet is an indicator that changes color over a range from PH = 0 to PH= 1.6. What is K(a) of methyl violet?
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i tried 10^-0.8 = 0.16
but thts not right .. can u plz chk ...
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Very close. Check the rounding and number of significant figures in your final answer.
its okay i got it it shld be 0.2
thnks
The pKa is 0.8
To calculate the Ka (acid dissociation constant) of methyl violet, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the concentrations of the acid and its conjugate base.
The equation is:
pH = pKa + log([A-]/[HA]),
where pH is the negative logarithm of the hydrogen ion concentration [H+], pKa is the negative logarithm of the Ka, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, methyl violet acts as an acid, and at the halfway point of color change (pH = 0.8), half of the indicator will be in the acid form and half in the conjugate base form.
Therefore, we can substitute the following values into the Henderson-Hasselbalch equation:
pH = 0.8,
[A-]/[HA] = 1 (since the acid and conjugate base concentrations are equal at the halfway point),
We can rearrange the equation to solve for pKa:
pKa = pH - log([A-]/[HA]).
Substituting the values:
pKa = 0.8 - log(1).
Since log(1) equals zero, the equation simplifies to:
pKa = 0.8.
To find Ka, we take the antilog (inverse logarithm) of pKa:
Ka = 10^pKa = 10^0.8.
Using a calculator, we find:
Ka ≈ 6.31
So the Ka of methyl violet is approximately 6.31.