Chapter 15: Problem 136
What is a base-10 logarithm?
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Chapter 15: Problem 136
What is a base-10 logarithm?
These are the key concepts you need to understand to accurately answer the question.
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When a strong acid is added to a buffer, the \(\mathrm{pH}\) changes a little bit. (a) Does the pH increase or decrease? (b) Why does the \(\mathrm{pH}\) change at all? Why doesn't the buffer hold the \(\mathrm{pH}\) constant?
It is possible to make two completely different buffers using the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (a) In one buffer, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) serves as the weak acid. What is the conjugate weak base? (b) Write the equations that show how the buffer in part (a) works when either \(\mathrm{H}_{3} \mathrm{O}^{+}\) or \(\mathrm{OH}^{-}\) is added. (c) In the other buffer, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) serves as the weak base. What is the conjugate weak acid? (d) Write the equations that show how the buffer in part (c) works when either \(\mathrm{H}_{3} \mathrm{O}^{+}\) or \(\mathrm{OH}^{-}\) is added.
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.
Use the Bronsted-Lowry definition to explain why \(\mathrm{NH}_{3}\) is a base in water.
Is a solution that contains \(0.10 \mathrm{M} \mathrm{HBr}\) and \(0.10 \mathrm{M}\) NaBr a buffered solution? Why or why not?
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