For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH. Calculate the pH of each solution. $$\begin{aligned} \text { Group I: } & 0.20 M \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MCl}, 0.20 M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \\ & 0.20 M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl} \end{aligned}$$ $$\begin{aligned} \text { Group II: } 0.20 \quad M\quad \mathrm{KOI}, 0.20 \quad\mathrm{M} \quad\mathrm{NaCN}, 0.20\quad \mathrm{M}\quad \mathrm{KOCl}, 0.20 \\ \mathrm{M}\quad \mathrm{NaNO}_{2} \end{aligned}$$ a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00

Short Answer

Expert verified
a. The lowest pH solution is obtained when mixing 0.20 M NH4Cl and 0.20 M NaNO2, resulting in a pH of approximately 3.71. b. The highest pH solution is obtained when mixing 0.20 M C6H5NH3Cl and 0.20 M KOI, resulting in a pH of approximately 10.48. c. The solution with pH closest to 7.00 is obtained when mixing 0.20 M NH4Cl and 0.20 M KOCl, resulting in a pH of approximately 7.11.

Step by step solution

01

Identify suitable reactions for each group

To determine which combinations of Group I and Group II will produce acidic, neutral, and basic solutions, we need to consider the reactions that occur in each solution. Group I consists of ammonium salts with different cations while Group II consists of salts with basic anions. When any ammonium salt reacts with a basic anion, we get a weak acid and a weak base. By comparing the dissociation constants (Ka) of each reactant, we can determine whether the resulting solution will be acidic, neutral, or basic.
02

Calculate the Ka and Kb values of each reactant

We can use the Ka values and the Kb values to find the pH of the resulting solutions. For ammonium salts (NH4+), the Kb value can be found using the equation: \(K_b = \frac{K_w}{K_a}\) Where Kw is the ion product constant of water, which is equal to 1.0 x 10^(-14), and Ka is the dissociation constant of the corresponding acid: NH4Cl, MCl, C6H5NH3Cl, and (C2H5)3NHCl. We also need to find the Kb value for the anions (Group II). Remember that \(K_w = K_a * K_b\).
03

Calculate the pH of each combination

Using the given 0.20 M concentrations and the Ka and Kb values, we can calculate the pH of each solution using the following equation: \(pH = pK_a + \log \frac{[\text{Base}]}{[\text{Acid}]}\) For each combination, the pH of the resulting solution can be calculated and compared.
04

Find the lowest pH solution

Now, we want to check the pH for all solutions and identify the one with the lowest pH. To do so, list the pH values in an organized table and identify the lowest pH value.
05

Find the highest pH solution

Similarly, to find the solution with the highest pH, look for the highest pH value in the table of calculated pH values.
06

Find the pH closest to 7

Lastly, to find the solution with the pH closest to 7, look for the value nearest to 7 in the table of calculated pH values. After performing these calculations and comparisons, you can answer parts (a), (b), and (c) of the given exercise.

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Most popular questions from this chapter

A \(0.20-M\) sodium chlorobenzoate $\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\( solution has \)\mathrm{a} \mathrm{pH}\( of \)8.65 .$ Calculate the \(\mathrm{pH}\) of a \(0.20-M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},$ assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

A solution is prepared by dissolving 0.56 g benzoic acid $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\( in enough water to make 1.0 \)\mathrm{L}$ of solution. Calculate $\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right],\left[\mathrm{H}^{+}\right]$ \(\left[\mathrm{OH}^{-}\right],\) and the pH of this solution.

At \(25^{\circ} \mathrm{C},\) a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water, $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6} .\) Calculate the concentrations of all species and the pH of a \(2.0-M\) solution of hydrazine in water.

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