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Chapter 17: Problem 26

The ionization constant \(K_{\mathrm{a}}\) of an indicator HIn is \(1.0 \times\) \(10^{-6}\). 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 \(\mathrm{pH}\) is \(4.00 ?\)

Short Answer

Expert verified
With the given pH of 4 and Ka value, calculate the ratio of the non-ionized to ionized form of the indicator; depending on which is more abundant, the observed color of the indicator would be red for the non-ionized form and yellow for the ionized form.

Step by step solution

01

Understand ionization equilibrium

The indicator \(HIn\) is subject to the following ionization equilibrium in aqueous medium: \[HIn \leftrightarrow H^{+} + In^{-}\]The color changes from red (non-ionized form, \(HIn\)) to yellow (ionized form, \(In^{-}\)) when \(HIn\) loses a proton H+ to become \(In^{-}\). It is important to note that pH affects this equilibrium, as pH is a measure of the concentration of \(H^{+}\) ions.
02

Calculate the hydrogen ion concentration

Based on the given pH value, the \(H^{+}\) ion concentration can be calculated using the formula: \[[H^{+}] = 10^{-\mathrm{pH}}\]The pH of the solution is 4. Therefore, the concentration of the hydrogen ion \([H^{+}]\) is \(10^{-4} M\).
03

Calculate the ionized to non-ionized form ratio

The ionization constant (Ka) represents the equilibrium between the ionized and non-ionized forms, which is given by \[K_{a} = \frac{[H^{+}][In^{-}]}{[HIn]}\]We know that \([H^{+}]\) = \([In^{-}]\) because for each \(HIn\) that ionizes, one \(H^{+}\) and one \(In^{-}\) are produced. So we replace in the Ka expression \([In^{-}]\) by \([H^{+}]\) which is equal to \(10^{-4}\), and we get\[K_{a} = ([H^{+}])^{2}/[HIn]\]From this it is possible to find the relation \([HIn]\)/\([H^{+}]\) = \(([H^{+}])/K_{a}\). As the value of Ka is given \((1.0 \times 10^{-6})\), the ratio of \([HIn]\) to \([H^{+}]\) in the solution can be calculated.
04

Determine the dominant form and color of the indicator

If the ratio obtained in the previous step is greater than 1, it means that the concentration of \(HIn\) (red) is higher than \(In^{-}\) (yellow), so the indicator will be red. If the ratio is less than 1, \(In^{-}\) (yellow) is more abundant, so the indicator will be yellow.

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Most popular questions from this chapter

For which of these reactions is the equilibrium constant called a solubility product? (a) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Zn}(\mathrm{OH})_{4}^{2-}(a q)\) (b) \(3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) \(\mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)

Both \(\mathrm{Ag}^{+}\) and \(\mathrm{Zn}^{2+}\) form complex ions with \(\mathrm{NH}_{3}\). Write balanced equations for the reactions. However, \(\mathrm{Zn}(\mathrm{OH})_{2}\) is soluble in \(6 \mathrm{M} \mathrm{NaOH},\) and \(\mathrm{AgOH}\) is not. Explain.

A sample of \(0.96 \mathrm{~L}\) of \(\mathrm{HCl}\) at \(372 \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\) is bubbled into \(0.034 \mathrm{~L}\) of \(0.57 \mathrm{MH}_{3}\). What is the \(\mathrm{pH}\) of the resulting solution? Assume the volume of solution remains constant and that the \(\mathrm{HCl}\) is totally dissolved in the solution.

One way to distinguish a buffer solution with an acid solution is by dilution. (a) Consider a buffer solution made of \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.500 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COONa}\). Calculate its \(\mathrm{pH}\) and the \(\mathrm{pH}\) after it has been diluted 10 -fold. (b) Compare the result in (a) with the pHs of a \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) solution before and after it has been diluted 10 -fold

Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) a strong acid reacting with a strong base, (b) a strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right),\) (c) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a strong base, \((\mathrm{d})\) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a weak base \(\left(\mathrm{NH}_{3}\right)\) (Hint: Strong acids exist as \(\mathrm{H}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up the \(K_{\mathrm{a}}, K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\) values.)
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