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Chapter 17: Problem 27

A buffer solution contains \(0.10 \mathrm{~mol}\) of acetic acid and $0.13 \mathrm{~mol}\( of sodium acetate in \)1.00 \mathrm{~L}$. (a) What is the \(\mathrm{pH}\) of this buffer? (b) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02\) mol of KOH? (c) What is the pH of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) ?

Short Answer

Expert verified
(a) The initial pH of the buffer solution is calculated using the Henderson-Hasselbalch equation, resulting in: \[pH = 4.75 + \log{\frac{0.13}{0.10}} = 4.97\] (b) After adding \(0.02~mol\) of KOH, the pH of the buffer solution is: \[pH = 4.75 + \log{\frac{0.15}{0.08}} = 5.30\] (c) After adding \(0.02~mol\) of HNO3, the pH of the buffer solution is: \[pH = 4.75 + \log{\frac{0.13}{0.10}} = 4.97\]

Step by step solution

01

Calculate the initial pH of the buffer solution

To calculate the pH of the buffer, we'll use the Henderson-Hasselbalch equation: \[pH = pK_a + \log{\frac{[A^-]}{[HA]}}\] Here, \(pK_a\) represents the negative logarithm of the acid dissociation constant of acetic acid (\(K_a\)), which is \(4.75\). \([A^-]\) and \([HA]\) are the concentrations of the conjugate base (acetate ion, coming from sodium acetate) and the weak acid (acetic acid), respectively. By substituting the given values, we get: \[pH = 4.75 + \log{\frac{0.13}{0.10}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\).
02

Calculate the pH after adding KOH

First, we need to find the amount of acetic acid and acetate ion after the reaction with KOH. The equation for the reaction between acetic acid and KOH is: \(CH_3COOH + KOH \rightarrow CH_3COOK + H_2O\) KOH reacts completely with acetic acid as it is a strong base. Thus, \(0.02~mol\) of acetic acid will react with \(0.02~mol\) of KOH, producing \(0.02~mol\) of the acetate ion. Update the molar amounts of acetic acid and acetate ion: Moles of acetic acid: \(0.10 - 0.02 = 0.08~mol\) Moles of acetate ion: \(0.13 + 0.02 = 0.15~mol\) Now, calculate the pH using the Henderson-Hasselbalch equation: \[pH = 4.75 + \log{\frac{0.15}{0.08}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\).
03

Calculate the pH after adding HNO3

Now, we need to find the amount of acetic acid and acetate ion after the reaction with HNO3. The equation for the reaction between the acetate ion and HNO3 is: \(CH_3COO^- + HNO_3 \rightarrow CH_3COOH + NO_3^-\) Since HNO3 is a strong acid, it reacts completely with acetate ions. Thus, \(0.02~mol\) of the acetate ion will react with \(0.02~mol\) of HNO3, producing \(0.02~mol\) of acetic acid. Update the molar amounts of acetic acid and acetate ion: Moles of acetic acid: \(0.08 + 0.02 = 0.10~mol\) (amount after step 2 + additional acetic acid) Moles of acetate ion: \(0.15 - 0.02 = 0.13~mol\) Now, calculate the pH using the Henderson-Hasselbalch equation: \[pH = 4.75 + \log{\frac{0.13}{0.10}}\] Calculate the value within the logarithm function and then add it to the \(pK_a\). This will give us the final pH after adding HNO3.

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Most popular questions from this chapter

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25{ }^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathrm{a})\) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution; (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{MHC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\) of \(0.22 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL} ;\) (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{Lof} 0.0850 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\)

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce \(\mathrm{a}\) buffer with \(\mathrm{pH}=3.75 ?\)

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3}\), (b) \(\mathrm{ZnS},(\mathrm{c}) \mathrm{BiI}_{3}\) (d) \(\mathrm{AgCN}\), (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\)

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) and \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to enough water to form \(2.00 \mathrm{~L}\) of solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5\) ? (b) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{-8}\).)
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