Chapter 17: Problem 81
Derive an equation similar to the Henderson-Hasselbalch equation relating the \(\mathrm{pOH}\) of a buffer to the \(\mathrm{p} K_{b}\) of its base component.
Chapter 17: Problem 81
Derive an equation similar to the Henderson-Hasselbalch equation relating the \(\mathrm{pOH}\) of a buffer to the \(\mathrm{p} K_{b}\) of its base component.
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Get started for free\(\mathrm{~A} 1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is slowly added to \(10.0 \mathrm{~mL}\) of a solution that is \(0.20 \mathrm{M}\) in \(\mathrm{Ca}^{2+}\) and \(0.30 \mathrm{M}\) in \(\mathrm{Ag}^{+}\). (a) Which compound will precipitate first: \(\mathrm{CaSO}_{4}\left(K_{s p}=2.4 \times 10^{-5}\right)\) or \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})\) How much \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation?
Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\) \(\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (Ignore any volume change.)
The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the \(\mathrm{pH}\) is \(4.68 .\) What is the \(\mathrm{p} K_{a}\) for bromcresol green?
(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of \(\mathrm{pH} 7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood \(\mathrm{pH}\) is \(7.1 ?\)
Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\).
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