Chapter 15: Problem 176
A buffer works by replacing added strong acid with weak acid. Explain how.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 15: Problem 176
A buffer works by replacing added strong acid with weak acid. Explain how.
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeWrite a chemical equation for the reaction between each pair of reactants, using single or double arrows as appropriate: (a) \(\mathrm{HNO}_{3}\) and \(\mathrm{OH}^{-}\) (b) \(\mathrm{HF}\) and \(\mathrm{OH}^{-}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\)
\(\mathrm{HZ}\) is a weak acid. An aqueous solution of \(\mathrm{HZ}\) is prepared by dissolving \(0.020\) mol of \(\mathrm{HZ}\) in sufficient water to yield \(1.0 \mathrm{~L}\) of solution. The \(\mathrm{pH}\) of the solution was \(4.93\) at \(25.0{ }^{\circ} \mathrm{C}\). What is the \(K_{\mathrm{a}}\) of \(\mathrm{HZ}\) ?
Amines are organic compounds that contain an \(\mathrm{NH}_{2}\) group, and water-soluble amines are weak bases in water. For example, the compound methylamine, \(\mathrm{H}_{3} \mathrm{C}-\mathrm{NH}_{2}\), is a weak base. (a) Draw a dot diagram for methylamine. (b) Using dot diagrams, show the equilibrium reaction between methylamine and water. (c) To which side does the equilibrium in part (b) lie? What did we tell you that allowed you to figure out the answer? (d) The similar compound ethane, \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\), does not act as a weak base. Why can methylamine act as a weak base but \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\) can't? (Hint: Draw a dot diagram for \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\).) (e) Is it appropriate to call methylamine an electrolyte? If so, is it weak or strong? Explain.
What is the \(\mathrm{pH}\) of a solution whose hydronium ion concentration is \(0.0010 \mathrm{M?}\) Is the solution acidic or basic?
A solution is prepared by dissolving \(2.50\) moles of \(\mathrm{LiOH}\) in enough water to get \(4.00 \mathrm{~L}\) of solution. What are the \(\mathrm{OH}^{-}\) and the \(\mathrm{H}_{3} \mathrm{O}^{+}\) molar concentrations?
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