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A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
The buffer made by dissolving NaHCO3 and Na2CO3 in water neutralizes added H+ and OH- ions through the following reactions: 1. Reaction of bicarbonate ion (HCO3-) with H+ ions: \( HCO_3^-(aq) + H^+(aq) \rightarrow H_2CO_3(aq) \) 2. Reaction of carbonate ion (CO3^2-) with H+ ions: \( CO_3^{2-}(aq) + H^+(aq) \rightarrow HCO_3^-(aq) \) 3. Reaction of bicarbonate ion (HCO3-) with OH- ions: \( HCO_3^-(aq) + OH^-(aq) \rightarrow H_2O(l) + CO_3^{2-}(aq) \)

Step by step solution

01

Identify the buffer components when they are dissolved in water

When NaHCO3 and Na2CO3 are dissolved in water, they will dissociate and form their respective ions: 1. \( NaHCO_3(s) \rightarrow Na^+(aq) + HCO_3^-(aq) \) 2. \( Na_2CO_3(s) \rightarrow 2Na^+(aq) + CO_3^{2-}(aq) \)
02

Write the equation for the reaction of bicarbonate ion with H+ ions

The bicarbonate ion (HCO3-) will react with the H+ ion, neutralizing it and forming carbonic acid (H2CO3). We can represent this reaction as follows: \( HCO_3^-(aq) + H^+(aq) \rightarrow H_2CO_3(aq) \)
03

Write the equation for the reaction of carbonate ion with H+ ions

The carbonate ion (CO3^2-) will also react with the H+ ion, neutralizing it and forming bicarbonate ion (HCO3-). We can represent this reaction as follows: \( CO_3^{2-}(aq) + H^+(aq) \rightarrow HCO_3^-(aq) \)
04

Write the equation for the reaction of bicarbonate ion with OH- ions

The bicarbonate ion (HCO3-) can also react with the added OH- ion. In this reaction, the bicarbonate ion (HCO3-) will donate a proton (H+) to the OH- ion, forming water (H2O) and the carbonate ion (CO3^2-). We can represent this reaction as follows: \( HCO_3^-(aq) + OH^-(aq) \rightarrow H_2O(l) + CO_3^{2-}(aq) \) With these four equations, we can now demonstrate how the given buffer consisting of NaHCO3 and Na2CO3 neutralizes added H+ and OH- ions through the reactions with bicarbonate and carbonate ions in the buffer system.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) HONH \(_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 M\) HONH \(_{3}\) Cl c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing 0.100 \(M \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

A buffer is prepared by dissolving \(\mathrm{HONH}_{2}\) and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)

Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

An aqueous solution contains dissolved \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} .\) The concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) is \(0.50 M\) and \(\mathrm{pH}\) is 4.20 a. Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) in this buffer solution. b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{L}\) of this solution. (Neglect any volume change.)

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