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A certain indicator HIn has a \(\mathrm{p} K_{\mathrm{a}}\) of 3.00 and a color change becomes visible when 7.00\(\%\) of the indicator has been converted to \(\mathrm{In}^{-}\) . At what pH is this color change visible?

Short Answer

Expert verified
The color change of the indicator HIn becomes visible at a pH of 1.88. This is calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), with a ratio of 7:93 for the converted form (In-) to the unconverted form (HIn), which gives a decimal ratio of 0.0753.

Step by step solution

01

Determine the ratio between the concentrations of In- and HIn

We know that 7.00% of the indicator has been converted to In-. This means that the ratio of the converted form (In-) to the unconverted form (HIn) is 7:93.
02

Convert the ratio to decimals

To make it easier to work with, let's convert the ratio into decimals: \[ \frac{[In^-]}{[HIn]} = \frac{7}{93} = 0.0753 \]
03

Use the Henderson-Hasselbalch equation to find the pH

We can now substitute the values into the Henderson-Hasselbalch equation to find the pH. pH = pKa + log([In-]/[HIn]) pH = 3.00 + log(0.0753) pH ≈ 3.00 + (-1.12) = 1.88
04

Conclusion

The color change of the indicator HIn becomes visible at a pH of 1.88.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
Calculating pH is a fundamental aspect of understanding acids and bases. It's a measure of how acidic or basic a solution is. The pH scale ranges from 0 to 14, with lower values being more acidic, higher values more basic, and 7 being neutral.
To calculate pH, especially when dealing with a weak acid and its conjugate base, the Henderson-Hasselbalch equation is often used:

\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]
  • \(pK_a\) is the acid dissociation constant, which tells us the strength of the acid.
  • \([A^-]\) is the concentration of the base form.
  • \([HA]\) is the concentration of the acid form.
In practice, this means comparing the concentration of the dissociated (base) form to the undissociated (acid) form. A higher ratio of base to acid increases pH, making the solution more basic. Calculating pH this way is crucial for predicting how an indicator will behave in different environments.
Acid-Base Indicators
Acid-base indicators are substances that change color depending on the pH of the solution they're in. This color change occurs because the indicator itself exists in two forms: an acid form \([HIn]\) and a base form \([In^-]\). Each form has a different color.
These indicators are helpful for visually determining the acidity or basicity of a solution without sophisticated equipment.

Here's how an indicator works:
  • The indicator's \(pK_a\) value is crucial, as it determines the pH range at which the color change is noticeable.
  • A typical range is usually around \(\pm 1\) pH unit from its \(pK_a\).
  • When a small percentage, such as 7% in this exercise, converts from \([HIn]\) to \([In^-]\), the color change can be visible. This indicates the pH value where the indicator begins to visually react.
The exact point where the color change occurs is dependent on the specific indicator and its chemical structure.
Chemical Equilibrium
Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time. For acid-base reactions, this implies a balance between the acid and its conjugate base.
The Henderson-Hasselbalch equation leverages this concept of equilibrium by relating it to pH calculation.

  • At equilibrium, the reaction isn't stopped but occurs at rates allowing the concentrations to be stable.
  • This balance is why specific ratios of \([In^-]\) and \([HIn]\) result in predictable pH values.
  • The exercise illustrates this with a 7:93 ratio, indicating a slight shift towards the base form \([In^-]\).
Understanding chemical equilibrium helps in predicting how a system will behave when conditions like pH or concentration change. It's pivotal for mastering more complex chemical reactions.

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Most popular questions from this chapter

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HONH}_{2}\) by 0.100 \(\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.

Carbonate buffers are important in regulating the pH of blood at \(7.40 .\) If the carbonic acid concentration in a sample of blood is 0.0012 M, determine the bicarbonate ion concentration required to buffer the pH of blood at pH \(=7.40\) $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7} $$

A best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

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