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Chapter 15: Problem 78

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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Most popular questions from this chapter

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) pyridine with 0.100\(M\) hydrochloric acid \(\left(K_{\mathrm{b}} \text { for pyridine is } 1.7 \times 10^{-9}\right) .\) Do not calculate the points at 24.9 and 25.1 \(\mathrm{mL}\)

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