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Chapter 14: Problem 30

A buffer is made up of \(355 \mathrm{~mL}\) each of \(0.200 \mathrm{M} \mathrm{NaHCO}_{3}\) and \(0.134 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Assuming that volumes are additive, calculate (a) the \(\mathrm{pH}\) of the buffer. (b) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0300 \mathrm{~mol}\) of \(\mathrm{HCl}\) to \(0.710 \mathrm{~L}\) of buffer. (c) the \(\mathrm{pH}\) of the buffer after the addition of \(0.0300 \mathrm{~mol}\) of \(\mathrm{KOH}\) to \(0.710 \mathrm{~L}\) of buffer.

Short Answer

Expert verified
Answer: The pH of the buffer after the addition of HCl is 8.66, and the pH after the addition of KOH is 11.23.

Step by step solution

01

Calculate initial concentrations of HCO3- and CO32-

To find the initial concentrations of HCO3- and CO32- in the buffer, we need to multiply the molarity of each by the volume of the solution. Since the volumes are additive, the total volume of the solution is doubled to 710 mL. Initial moles of HCO3-: \(0.200 \mathrm{M} \times 355 \mathrm{~mL} = 0.0710 \mathrm{~mol}\) Initial moles of CO32-: \(0.134 \mathrm{M} \times 355 \mathrm{~mL} = 0.0475 \mathrm{~mol}\) Initial concentrations of HCO3- and CO32- in the buffer: \([\mathrm{HCO}_{3}^{-}] = \frac{0.0710 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.100 \mathrm{M}\) \([\mathrm{CO}_{3}^{2-}] = \frac{0.0475 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.0669 \mathrm{M}\)
02

Calculate the initial pH of the buffer

Now we can use the Henderson-Hasselbalch equation to find the initial pH: \(\mathrm{pH} = \mathrm{p}K_{a} + \log\left(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\) For the bicarbonate/carbonate buffer system, we can use the second ionization constant for carbonic acid, with \(pK_{a2} = 10.32\). Using this value and our initial concentrations, we can calculate the pH: \(\mathrm{pH} = 10.32 + \log\left(\frac{0.0669 \mathrm{M}}{0.100 \mathrm{M}}\right)\) \(\mathrm{pH} = 10.32 - 0.283\) \(\mathrm{pH} = 10.04\) So the initial pH of the buffer is 10.04.
03

Calculate the new concentrations after the addition of HCl

Now let's calculate the new concentrations of HCO3- and CO32- after the addition of 0.0300 mol of HCl. HCl reacts with CO32- to form HCO3-, so moles of HCO3- will increase and moles of CO32- will decrease: New moles of HCO3-: \(0.0710 \mathrm{~mol} + 0.0300 \mathrm{~mol} = 0.101 \mathrm{~mol}\) New moles of CO32-: \(0.0475 \mathrm{~mol} - 0.0300 \mathrm{~mol} = 0.0175 \mathrm{~mol}\) New concentrations of HCO3- and CO32- after the addition of HCl: \([\mathrm{HCO}_{3}^{-}] = \frac{0.101 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.142 \mathrm{M}\) \([\mathrm{CO}_{3}^{2-}] = \frac{0.0175 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.0246 \mathrm{M}\)
04

Calculate pH after the addition of HCl

Now we can use the new concentrations and the Henderson-Hasselbalch equation to calculate the pH after the addition of HCl: \(\mathrm{pH} = 10.32 + \log\left(\frac{0.0246 \mathrm{M}}{0.142 \mathrm{M}}\right)\) \(\mathrm{pH} = 10.32 - 1.66\) \(\mathrm{pH} = 8.66\) So the pH of the buffer after the addition of HCl is 8.66.
05

Calculate the new concentrations after the addition of KOH

Now let's calculate the new concentrations of HCO3- and CO32- after the addition of 0.0300 mol of KOH. KOH reacts with HCO3- to form CO32-, so moles of HCO3- will decrease and moles of CO32- will increase: New moles of HCO3-: \(0.0710 \mathrm{~mol} - 0.0300 \mathrm{~mol} = 0.0410 \mathrm{~mol}\) New moles of CO32-: \(0.0475 \mathrm{~mol} + 0.0300 \mathrm{~mol} = 0.0775 \mathrm{~mol}\) New concentrations of HCO3- and CO32- after the addition of KOH: \([\mathrm{HCO}_{3}^{-}] = \frac{0.0410 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.0577 \mathrm{M}\) \([\mathrm{CO}_{3}^{2-}] = \frac{0.0775 \mathrm{~mol}}{0.710 \mathrm{~L}} = 0.109 \mathrm{M}\)
06

Calculate pH after the addition of KOH

Finally, we can use the new concentrations and the Henderson-Hasselbalch equation to calculate the pH after the addition of KOH: \(\mathrm{pH} = 10.32 + \log\left(\frac{0.109 \mathrm{M}}{0.0577 \mathrm{M}}\right)\) \(\mathrm{pH} = 10.32 + 0.907\) \(\mathrm{pH} = 11.23\) So the pH of the buffer after the addition of KOH is 11.23. To summarize: (a) The initial pH of the buffer is 10.04. (b) The pH of the buffer after the addition of HCl is 8.66. (c) The pH of the buffer after the addition of KOH is 11.23.

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Most popular questions from this chapter

Phenol red is an indicator with a \(\mathrm{pK}_{\mathrm{a}}\) of \(7.4\). It is yellow in acid solution and red in alkaline. (a) What is its \(K_{\mathrm{a}}\) ? (b) What is its \(\mathrm{pH}\) range? (c) What would its color be at \(\mathrm{pH}\) 7.4?

Explain why it is not possible to prepare a buffer with a pH of \(6.50\) by mixing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\).

Morphine, \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{O}_{3} \mathrm{~N}\), is a weak base \(\left(K_{\mathrm{b}}=7.4 \times 10^{-7}\right) .\) Consider its titration with hydrochloric acid. In the titration, \(50.0 \mathrm{~mL}\) of a \(0.1500 \mathrm{M}\) solution of morphine is titrated with \(0.1045 \mathrm{MHCl}\). (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of hydrochloric acid is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{HCl}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

WEB Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{OH}^{-}\) ions. (a) ammonium nitrate (b) sodium dihydrogen phosphate \(\left(\mathrm{NaH}_{2} \mathrm{PO}_{4}\right)\) (c) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\)
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