A buffer is prepared by adding 20.0 g of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to 500 \(\mathrm{mL}\) of a 0.150 \(\mathrm{M}\) 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 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.

Short Answer

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a) The pH of the buffer is \(6.02\). b) The complete ionic equation for the reaction with hydrochloric acid is: \(\mathrm{H}^{+} (aq) + \mathrm{CH}_{3} \mathrm{COO}^{-} (aq) \rightarrow \mathrm{CH}_{3} \mathrm{COOH} (aq)\) c) The complete ionic equation for the reaction with sodium hydroxide is: \(\mathrm{CH}_{3} \mathrm{COOH} (aq) + \mathrm{OH}^{-} (aq) \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-} (aq) + \mathrm{H}_{2}\mathrm{O} (l)\)

Step by step solution

01

Calculation of moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\)

First, calculate the moles of acetic acid in the 500 mL \(0.150\ \mathrm{M}\) solution: moles of \(\mathrm{CH}_{3} \mathrm{COOH} = 0.150\ \mathrm{moles/L} \times 0.500\ \mathrm{L} = 0.075\ \mathrm{moles}\) Next, calculate the moles of sodium acetate by dividing the mass by the molar mass: moles of \(\mathrm{CH}_{3} \mathrm{COONa} = \frac{20.0\ \mathrm{g}}{\mathrm{molar\ mass\ of\ CH}_{3} \mathrm{COONa}}\) Molar mass of \(\mathrm{CH}_{3} \mathrm{COONa} = 82.03\ \mathrm{g/mol}\), so the moles are: moles of \(\mathrm{CH}_{3} \mathrm{COONa} = \frac{20.0\ \mathrm{g}}{82.03\ \mathrm{g/mol}} = 0.244\ \mathrm{moles}\)
02

Application of the Henderson-Hasselbalch equation

Now, we will use the Henderson-Hasselbalch equation to determine the pH of the buffer: pH \(= pK_a + \log{\frac{[\mathrm{CH}_{3} \mathrm{COO}^{-}]}{[\mathrm{CH}_{3} \mathrm{COOH}]}}\) Using the \(K_a\) of acetic acid, \(1.8 \times 10^{-5}\), we find the \(pK_a = -\log{1.8 \times 10^{-5}} = 4.74\). Now, we'll plug in the moles of \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) calculated in Step 1: pH \(= 4.74 + \log{\frac{0.244}{0.075}} = 4.74 + 1.28 = 6.02\) Thus, the pH of the buffer is 6.02.
03

Ionic equation for the reaction with hydrochloric acid

When a few drops of hydrochloric acid, \(\mathrm{HCl}\), are added to the buffer, it will react with the conjugate base, \(\mathrm{CH}_{3} \mathrm{COO}^{-}\). The complete ionic equation for the reaction is: \(\mathrm{H}^{+} (aq) + \mathrm{CH}_{3} \mathrm{COO}^{-} (aq) \rightarrow \mathrm{CH}_{3} \mathrm{COOH} (aq)\)
04

Ionic equation for the reaction with sodium hydroxide

When a few drops of sodium hydroxide, \(\mathrm{NaOH}\), are added to the buffer, it will react with the weak acid, \(\mathrm{CH}_{3} \mathrm{COOH}\). The complete ionic equation for the reaction is: \(\mathrm{CH}_{3} \mathrm{COOH} (aq) + \mathrm{OH}^{-} (aq) \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-} (aq) + \mathrm{H}_{2}\mathrm{O} (l)\)

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Most popular questions from this chapter

A solution contains three anions with the following concentrations: \(0.20 M \mathrm{CrO}_{4}^{2-}, 0.10 M \mathrm{CO}_{3}^{2-}\) , a n d 0.010\(M \mathrm{Cl}^{-} .\) If a dilute AgNO \(_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{s p}=1.2 \times 10^{-12}\right), \mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{s p}=8.1 \times 10^{-12}\right)\) or \(\mathrm{AgCl}\left(K_{s p}=1.8 \times 10^{-10}\right) ?\)

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

How many milliliters of 0.0850\(M \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) 40.0 \(\mathrm{mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3},\) (\mathbf{b} ) 35.0 \(\mathrm{mL}\) of \(0.0850 M \mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) 50.0 \mathrm{mL}\) of a solution that contains 1.85 \(\mathrm{g}\) of \(\mathrm{HCl}\) per liter?

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2} \mathrm{g} \mathrm{SrF}_{2}\) dissolves per 100 \(\mathrm{mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\) Calculate the solubility product for \(\mathrm{SrF}_{2 .}(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} ?\)

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