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Chapter 14: Problem 66

What are the major species present in 0.250\(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) b. HCN

Short Answer

Expert verified
For a 0.250 M solution of \(\mathrm{HOC}_{6} \mathrm{H}_{5}\), the major species present are the intact acid molecule (\(\mathrm{HOC}_{6} \mathrm{H}_{5}\)), its conjugate base (\(\mathrm{OC}_{6} \mathrm{H}_{5}^-\)), and water (\(\mathrm{H_2 O}\)) with hydroxide ion (\(\mathrm{OH}^-\)). Using the Ka value and equilibrium expression, we can calculate the pH of the solution. Similarly, for a 0.250 M solution of HCN, the major species present are the intact acid molecule (HCN), its conjugate base (\(\mathrm{CN}^-\)), and water (\(\mathrm{H_2 O}\)) with hydroxide ion (\(\mathrm{OH}^-\)). Using the Ka value and equilibrium expression, we can calculate the pH of the solution.

Step by step solution

01

Major Species

For a weak acid \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) in solution, we have: 1. The intact acid molecule (\(\mathrm{HOC}_{6} \mathrm{H}_{5}\)) 2. The conjugate base, after releasing a proton: \(\mathrm{OC}_{6} \mathrm{H}_{5}^{-}\) 3. Water (\(\mathrm{H_2 O}\)) and the hydroxide ion (\(\mathrm{OH}^-\)) from the auto-ionization of water. Step 2: Determine the equilibrium expression
02

Equilibrium Expression

The equilibrium expression for the ionization of \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) is: \[ \mathrm{HOC}_{6} \mathrm{H}_{5} \rightleftharpoons \mathrm{H}^+ + \mathrm{OC}_{6} \mathrm{H}_{5}^- \] The equilibrium constant Ka can be expressed as: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{OC}_{6} \mathrm{H}_{5}^-]}{[\mathrm{HOC}_{6} \mathrm{H}_{5}]} \] Step 3: Calculate the pH of 0.250 M \(\mathrm{HOC}_{6} \mathrm{H}_{5}\)
03

Calculate pH

Assuming the initial concentration of the acid as 0.250 M and negligible ionization of water, we will solve the Ka expression for [H+] and then use the pH formula: \[ \mathrm{pH} = -\log{[\mathrm{H}^+]} \] Using the Ka values for \(\mathrm{HOC}_{6} \mathrm{H}_{5}\), we can obtain the pH. #b. Calculation for HCN solution# Step 1: Identify the major species
04

Major Species

For a weak acid HCN in solution, we have: 1. The intact acid molecule (HCN) 2. The conjugate base, after releasing a proton: \(\mathrm{CN}^-\) 3. Water (\(\mathrm{H_2 O}\)) and the hydroxide ion (\(\mathrm{OH}^-\)) from the auto-ionization of water. Step 2: Determine the equilibrium expression
05

Equilibrium Expression

The equilibrium expression for the ionization of HCN is: \[ \mathrm{HCN} \rightleftharpoons \mathrm{H}^+ + \mathrm{CN}^- \] The equilibrium constant Ka can be expressed as: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{CN}^-]}{[\mathrm{HCN}]} \] Step 3: Calculate the pH of 0.250 M HCN
06

Calculate pH

Assuming the initial concentration of the acid as 0.250 M and negligible ionization of water, we will solve the Ka expression for [H+] and then use the pH formula: \[ \mathrm{pH} = -\log{[\mathrm{H}^+]} \] Using the Ka values for HCN, we can obtain the pH.

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Most popular questions from this chapter

Calculate the percent dissociation of the acid in each of the following solutions. a. 0.50\(M\) acetic acid b. 0.050\(M\) acetic acid c. 0.0050\(M\) acetic acid d. Use Le Châtelier's principle to explain why percent dissociation increases as the concentration of a weak acid decreases. e. Even though the percent dissociation increases from solutions a to \(c,\) the \(\left[\mathrm{H}^{+}\right]\) decreases. Explain.

The pH of human blood is steady at a value of approximately 7.4 owing to the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) $$ Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were $26.3 \mathrm{mM},\( whereas his \)\mathrm{CO}_{2}\( levels were 1.63 \)\mathrm{mM}$ . On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to 24.9 \(\mathrm{m} M\) . What was the \(\mathrm{CO}_{2}\) blood level at rest?

The pH of a \(0.063-M\) solution of hypobromous acid (HOBr but usually written \(\mathrm{HBrO}\) ) is \(4.95 .\) Calculate \(K_{\mathrm{a}} .\)

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a $0.10-M \mathrm{H}_{2} \mathrm{S}\( solution. Assume \)K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}$

What mass of \(\mathrm{NaOH}(s)\) must be added to 1.0 \(\mathrm{L}\) of 0.050 \(\mathrm{M}\) \(\mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than 0.0010\(\% ?\) Assume no volume change on addition of \(\mathrm{NaOH} .\)
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