Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. \(\mathrm{NH}_{3}\) b. KOH c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KCl}\) e. HCl

Short Answer

Expert verified
The solutions can be ranked in order of increasing pH as follows: HCl (strong acid), \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (weak acid), \(\mathrm{KCl}\) (neutral), \(\mathrm{NH}_{3}\) (weak base), and KOH (strong base).

Step by step solution

01

Identify the nature of each compound

We have five compounds: \(\mathrm{NH}_{3}\), KOH, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), \(\mathrm{KCl}\), and HCl. Let's identify their nature: a. \(\mathrm{NH}_{3}\) is a weak base b. KOH is a strong base c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is a weak acid d. \(\mathrm{KCl}\) is a salt of a strong acid and strong base, thus neutral e. HCl is a strong acid
02

Determine the pH for each solution

Now we will calculate the pH for each solution, considering their nature and concentration. a. Ammonia (\(\mathrm{NH}_{3}\)): As a weak base, we need to use a simplified version of the Henderson-Hasselbalch equation: \(\mathrm{pOH} = \mathrm{pK}_{b} - \log{\frac{[\mathrm{NH}_{3}]}{[\mathrm{OH}^-]}}\). However, for ranking purpose, we don't need the exact pH value. b. Potassium hydroxide (KOH): As a strong base, it fully dissociates in water: \(\mathrm{KOH} \rightarrow \mathrm{K}^+ + \mathrm{OH}^-\). Hence, \([\mathrm{OH}^-]\) will be \(0.10\,\mathrm{M}\). We can calculate the pOH as \(\mathrm{pOH} = -\log{[\mathrm{OH}^-]}\), and then find the pH using \(\mathrm{pH} = 14 - \mathrm{pOH}\). c. Acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)): As a weak acid, we can also use the simplified version of the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{pK}_{a} - \log{\frac{[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}]}{[\mathrm{H}^+]}}\). Again, we don't need the exact pH value for ranking purposes. d. Potassium chloride (\(\mathrm{KCl}\)): As a neutral salt, the pH will be 7. e. Hydrochloric acid (HCl): As a strong acid, it fully dissociates in water: \(\mathrm{HCl} \rightarrow \mathrm{H}^+ + \mathrm{Cl}^-\). Hence, \([\mathrm{H}^+]\) will be \(0.10\,\mathrm{M}\). We can calculate the pH as \(\mathrm{pH} = -\log{[\mathrm{H}^+]}\).
03

Rank the solutions in order of increasing pH

Based on the pH calculations and the nature of each compound, we can rank the solutions as follows: 1. Hydrochloric acid (HCl): strong acid 2. Acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)): weak acid 3. Potassium chloride (\(\mathrm{KCl}\)): neutral 4. Ammonia (\(\mathrm{NH}_{3}\)): weak base 5. Potassium hydroxide (KOH): strong base

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Most popular questions from this chapter

One mole of a weak acid HA was dissolved in 2.0 \(\mathrm{L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 \(\mathrm{M} .\) Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HA}\) .

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.

A solution is made by adding 50.0 \(\mathrm{mL}\) of 0.200\(M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to 50.0 \(\mathrm{mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the pH of the solution. b. Calculate the acetate ion concentration.

Will 0.10 M solutions of the following salts be acidic, basic, or neutral? See Appendix 5 for \(K_{\mathrm{a}}\) values. a. ammonium bicarbonate b. sodium dihydrogen phosphate c. sodium hydrogen phosphate d. ammonium dihydrogen phosphate e. ammonium formate

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between $\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\( and \)\left[\mathrm{CO}_{3}^{2-}\right] ?$ c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

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