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Chapter 17: Problem 23

A buffer is prepared by adding \(15.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

Short Answer

Expert verified
(a) The pH of the buffer is 6.10. (b) The net ionic equation for the reaction with nitric acid is: \(CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)\). (c) The net ionic equation for the reaction with potassium hydroxide is: \(CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)\).

Step by step solution

01

(a) Calculate the amount of sodium acetate)

First, we need to find the amount of sodium acetate in moles. To do that, we can use the formula: moles = mass / molar mass. The molar mass of sodium acetate (CH3COONa) is: \(12.01g/mol(C) * 2 + 1.01g/mol(H) * 3 + 15.99g/mol(O) * 2 + 22.99g/mol(Na) = 82.03g/mol\) Given mass of sodium acetate = \(15.0 g\) So, moles of sodium acetate = \(15.0g / 82.03 g/mol = 0.183 mol\)
02

(a) Calculate the concentration of sodium acetate)

Now, we calculate the concentration of sodium acetate by dividing the moles by the volume of the solution. Given volume of acetic acid solution = 500 mL Concentration of sodium acetate = \(\frac{0.183 mol}{0.500 L} = 0.366 M\)
03

(a) Find the pKa of acetic acid)

We know that the \(K_a\) value for acetic acid is \(1.8 * 10 ^{-5}\). We can calculate the pKa value using the formula: pKa = -log \(K_a\). pKa = -log(1.8 * 10 ^{-5}) = 4.74
04

(a) Determine the pH of the buffer)

Now we can calculate the pH of the buffer using the Henderson-Hasselbalch equation: pH = pKa + log \(\frac{[CH_3COO^-]}{[CH_3COOH]}\) pH = 4.74 + log \(\frac{0.366}{0.100}\) pH = 4.74 + 1.36 = \(6.10\) So, the pH of the buffer is 6.10.
05

(b) Reaction with nitric acid)

When nitric acid (HNO3) is added to the buffer, it reacts with the acetate ions: \(CH_3COO^-(aq) + HNO_3(aq) \rightarrow CH_3COOH(aq) + NO_3^-(aq)\)
06

(b) Complete ionic equation with nitric acid)

The complete ionic equation for the reaction above is: \(CH_3COO^-(aq) + H^+(aq) + NO_3^-(aq) \rightarrow CH_3COOH(aq) + NO_3^-(aq)\) NO3- ions are not involved in the reaction, so they can be considered as spectator ions. We can remove them to get the net ionic equation: \(CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)\)
07

(c) Reaction with potassium hydroxide)

When potassium hydroxide (KOH) is added to the buffer, it reacts with the acetic acid: \(CH_3COOH(aq) + KOH(aq) \rightarrow CH_3COOK(aq) + H_2O(l)\)
08

(c) Complete ionic equation with potassium hydroxide)

The complete ionic equation for the reaction above is: \(CH_3COOH(aq) + K^+(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + K^+(aq) + H_2O(l)\) K+ ions are not involved in the reaction, so they can be considered as spectator ions. We can remove them to get the net ionic equation: \(CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)\)

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Most popular questions from this chapter

Consider the titration of \(30.0 \mathrm{~mL}\) of $0.050 \mathrm{M} \mathrm{NH}_{3}\( with \)0.025 M$ HCl. Calculate the pH after the following volumes of titrant have been added: $(\mathbf{a}) 0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},\( (c) 59.0 \)\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .$

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be 4.20 . What is the \(\mathrm{p} K_{a}\) for the unknown acid?

You have to prepare a \(\mathrm{pH}=2.50\) buffer, and you have the following \(0.20 \mathrm{M}\) solutions available: $\mathrm{HCOOH}, \mathrm{CH}_{3} \mathrm{COOH},\( \)\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{KCH}_{3} \mathrm{COO}, \mathrm{KHCOO},\( and \)\mathrm{KH}_{2} \mathrm{PO}_{4} .$ Which solutions would you use? How many liters of each solution would you use to make approximately 2 L of the buffer?

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Aspirin has the structural formula At body temperature $\left(37^{\circ} \mathrm{C}\right), K_{a}\( for aspirin equals \)3 \times 10^{-5}$. If two aspirin tablets, each having a mass of \(325 \mathrm{mg}\), are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is 2 , what percent of the aspirin is in the form of neutral molecules?
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