The ionization constant Ka of an indicator HIn is 1.0x10^-6. The color of the nonionized form is red and that of the ionized form is yellow. What is the color of this indicator in a solution whose pH is 4.00? (The question mentions that you can use the ratio of HIn/In).
I NEED this step-by-step. Please don't just give me the answer, please explain how you got there and what you did. I have to find the concentrations of HIn, In and H, correct? I can get the concentration of H from pH, but that's all I can do.
To determine the color of the indicator in a solution with a pH of 4.00, we need to calculate the concentrations of HIn (nonionized form) and In (ionized form) in the solution.
Step 1: Write the balanced chemical equation for the ionization of HIn:
HIn ⇌ H⁺ + In⁻
Step 2: Define the ionization constant Ka:
Ka = [H⁺][In⁻] / [HIn]
Step 3: Apply the Ka expression to the given information:
Ka = 1.0x10^-6
Step 4: Set up an ICE table to determine the concentrations of HIn and In at equilibrium:
| HIn | H⁺ | In⁻
---------------------------------
Initial | ??? | 0 | 0
Change | -x | +x | +x
Equilibrium | ???-x | x | x
Step 5: Use the relationship between [HIn], [H⁺], and [In⁻] expressed in terms of x:
[HIn] = ??? - x
[H⁺] = x
[In⁻] = x
Step 6: Plug the equilibrium concentrations into the Ka expression:
Ka = [H⁺][In⁻] / [HIn]
1.0x10^-6 = x*x / (??? - x)
Step 7: Since the question mentions the ratio of HIn/In, divide both sides of the equation by x:
[H⁺]/[HIn] = x/(??? - x)
Step 8: Now, we can relate [H⁺] to the pH of the solution:
pH = -log[H⁺]
Therefore, [H⁺] = 10^(-pH)
Step 9: Substitute the pH value of 4 into the equation to find [H⁺]:
[H⁺] = 10^(-4) = 0.0001
Step 10: Substitute the values of [H⁺] and Ka into the equation from Step 7:
0.0001 / [HIn] = x / (??? - x)
Step 11: Assume that the dissociation of HIn is relatively small, so the concentration of HIn in the denominator is approximately equal to the initial concentration of HIn:
0.0001 / [HIn] ≈ x / [HIn]
Step 12: Simplify the equation:
0.0001 ≈ x
Step 13: Since x represents the concentration of [In⁻], we can conclude that the concentration of In⁻ is approximately 0.0001.
Step 14: As the color of the ionized form (In⁻) is yellow and the color of the nonionized form (HIn) is red, we can infer that the resulting indicator color in the solution with a pH of 4.00 is yellow.
To determine the color of the indicator in a solution with a pH of 4.00, we need to find the concentrations of HIn (nonionized form) and In (ionized form).
Step 1: Determine the concentration of H
The pH of a solution provides the concentration of H ions. In this case, the pH is 4.00, so the concentration of H ions is given by:
[H] = 10^(-pH) = 10^(-4.00) = 1.0 x 10^(-4) M
Step 2: Use the ionization constant (Ka) to find the concentration of HIn and In
Ka is given as 1.0 x 10^(-6), and we can use the ratio of HIn/In to find their concentrations.
Ka = [H][In] / [HIn]
Since the concentrations of H and [H] are equal (since [H] = [H]), the equation becomes:
Ka = [H]^2 / [HIn]
Now we can rearrange the equation and solve for the concentration of In:
[HIn] = [H]^2 / Ka
[HIn] = (1.0 x 10^(-4))^2 / (1.0 x 10^(-6))
[HIn] = 1.0 x 10^(-4) M
Step 3: Calculate the concentration of In
To calculate the concentration of In, we can subtract the concentration of HIn from the concentration of H:
[In] = [H] - [HIn]
[In] = 1.0 x 10^(-4) - 1.0 x 10^(-4)
[In] = 0
Step 4: Determine the color of the indicator
Based on the given information, the nonionized form of the indicator (HIn) is red, and the ionized form (In) is yellow.
Since the concentration of In is 0 (as calculated in step 3), there is no ionized form present in the solution. Therefore, the color of the indicator in this solution with a pH of 4.00 should be red.
pH = pKa + log (In^-)/(HIn)
Using pKa and pH of 4.00, you can solve for (In/HIn)
Remember the ratio must be at least 10:1 to change color (or 1:10 depending upon how you look at it).