Determining the Overall Acidity or Basicity of Salt Solutions

Determine if the solution formed by each salt is acidic, basic, or neutral:

(a) SrCl $_{2}$ (b) AlBr $_{3}$ (c) $CH_{3}NH_{3}NO_{3}$ (d) NaCHO2 (e) NH $_{4}$ F

Question

Determining the Overall Acidity or Basicity of Salt Solutions

Determine if the solution formed by each salt is acidic, basic, or neutral:

(a) SrCl[latex]_{2}[/latex] (b) AlBr[latex]_{3}[/latex] (c) [latex]CH_{3}NH_{3}NO_{3}[/latex] (d) NaCHO2 (e) NH[latex]_{4}[/latex]F

Accepted Answer

(a) The Sr2+ cation is the counterion of a strong base [Sr(OH)[latex]_{2}[/latex] ]and is therefore, pH-neutral. The Cl[latex]^{-}[/latex] anion is the conjugate base of a strong acid (HCl) and is therefore, pH-neutral as well. The SrCl[latex]_{2}[/latex] solution is therefore, pH-neutral (neither acidic nor basic).
(b) The Al[latex]^{3+}[/latex] cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) and is therefore, a weak acid. The Br[latex]^{-}[/latex] anion is the conjugate base of a strong acid (HBr) and is therefore, pH-neutral. The AlBr[latex]_{3}[/latex] solution is therefore, acidic.
(c) The [latex]CH_{3}NH_{3}^{+}[/latex] ion is the conjugate acid of a weak base ([latex]CH_{3}NH_{2}[/latex]) and is therefore, acidic. The NO[latex]_{3}^{-}[/latex] anion is the conju gate base of a strong acid (HNO[latex]_{3}[/latex]) and is therefore, pH-neutral. The [latex]CH_{3}NH_{3}NO_{3}[/latex] solution is therefore, acidic.
(d) The Na[latex]^{+}[/latex] cation is the counterion of a strong base and is there fore, pH-neutral. The CHO[latex]_{2}^{-}[/latex] anion is the conjugate base of a weak acid and is therefore, basic. The NaCHO[latex]_{2}[/latex] solution is therefore, basic.
e) The NH[latex]_{4}^{+}[/latex] ion is the conjugate acid of a weak base (NH[latex]_{3}[/latex]) and is therefore, acidic. The F[latex]^{-}[/latex] ion is the conjugate base of a weak acid and is therefore, basic. To determine the overall acidity or basicity of the solution, compare the values of K[latex]_{a}[/latex] for the acidic cation and K[latex]_{b}[/latex] for the basic anion. Obtain each value of K from the conjugate by using [latex]K_{a} imes K_{b} = K_{w}[/latex]. Since K[latex]_{a}[/latex] is greater than K[latex]_{b}[/latex], the solution is acidic.	K[latex]_{a}(NH_{4}^{+}) = \frac{K_{w}}{K_{b}(NH_{3}) }= \frac{1.0 imes 10^{-14}}{1.76 imes 10^{-5}}[/latex] = 5.68 × 10[latex]^{-10}[/latex] K[latex]_{b}(F^{-}) = \frac{K_{w}}{K_{a}(HF)} = \frac{1.0 imes 10^{-14}}{3.5 imes 10^{-4}}[/latex] = 2.9 × 10[latex]^{-11}[/latex] K[latex]_{a}[/latex] > K[latex]_{b}[/latex]

(a) The Sr2+ cation is the counterion of a strong base [Sr(OH) $_{2}$ ]and is therefore, pH-neutral. The Cl $^{-}$ anion is the conjugate base of a strong acid (HCl) and is therefore, pH-neutral as well. The SrCl $_{2}$ solution is therefore, pH-neutral (neither acidic nor basic).
(b) The Al $^{3+}$ cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) and is therefore, a weak acid. The Br $^{-}$ anion is the conjugate base of a strong acid (HBr) and is therefore, pH-neutral. The AlBr $_{3}$ solution is therefore, acidic.
(c) The $CH_{3}NH_{3}^{+}$ ion is the conjugate acid of a weak base ( $CH_{3}NH_{2}$ ) and is therefore, acidic. The NO $_{3}^{-}$ anion is the conju gate base of a strong acid (HNO $_{3}$ ) and is therefore, pH-neutral. The $CH_{3}NH_{3}NO_{3}$ solution is therefore, acidic.
(d) The Na $^{+}$ cation is the counterion of a strong base and is there fore, pH-neutral. The CHO $_{2}^{-}$ anion is the conjugate base of a weak acid and is therefore, basic. The NaCHO $_{2}$ solution is therefore, basic.
e) The NH $_{4}^{+}$ ion is the conjugate acid of a weak base (NH $_{3}$ ) and is therefore, acidic. The F $^{-}$ ion is the conjugate base of a weak acid and is therefore, basic. To determine the overall acidity or basicity of the solution, compare the values of K $_{a}$ for the acidic cation and K $_{b}$ for the basic anion. Obtain each value of K from the conjugate by using $K_{a} \times K_{b} = K_{w}$ . Since K $_{a}$ is greater than K $_{b}$ , the solution is acidic.	K $_{a}(NH_{4}^{+}) = \frac{K_{w}}{K_{b}(NH_{3}) }= \frac{1.0 \times 10^{-14}}{1.76 \times 10^{-5}}$ = 5.68 × 10 $^{-10}$ K $_{b}(F^{-}) = \frac{K_{w}}{K_{a}(HF)} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}}$ = 2.9 × 10 $^{-11}$ K $_{a}$ > K $_{b}$

Question 15.15: Determining the Overall Acidity or Basicity of Salt Solution...

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