Identify the following compounds as strongly acidic, weakly acidic, weakly basic, strongly basic, or pH-neutral in water at $25^{\circ} C$ $25^@ "C"$ ? $Al {({NO}_{3})}_{3}$ $"Al"("NO"_3)_3$ , $C_{2} H_{5} {NH}_{3} {NO}_{3}$ $"C"_2"H"_5"NH"_3"NO"_3$ , $NaClO$ $"NaClO"$ , $KCl$ $"KCl"$ , $C_{2} H_{5} {NH}_{3} CN$ $"C"_2"H"_5"NH"_3"CN"$

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Identify the following compounds as strongly acidic, weakly acidic, weakly basic, strongly basic, or pH-neutral in water at $25^{\circ} C$ $25^@ "C"$ ? $Al {({NO}_{3})}_{3}$ $"Al"("NO"_3)_3$ , $C_{2} H_{5} {NH}_{3} {NO}_{3}$ $"C"_2"H"_5"NH"_3"NO"_3$ , $NaClO$ $"NaClO"$ , $KCl$ $"KCl"$ , $C_{2} H_{5} {NH}_{3} CN$ $"C"_2"H"_5"NH"_3"CN"$

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Answer 1 · 2017-11-03T04:46:16

These are mostly straightforward, except for one.

$Al {({NO}_{3})}_{3}$ $"Al"("NO"_3)_3$ is strongly acidic because ${Al}^{3 +}$ $"Al"^(3+)$ is a Lewis acid. It has an empty $3 p_{z}$ $3p_z$ orbital that accepts electron density, making it an electron pair acceptor, a Lewis acid.
$C_{2} H_{5} {NH}_{3} {NO}_{3}$ $"C"_2"H"_5"NH"_3"NO"_3$ is weakly acidic because the cation is the conjugate acid of the weak base ethylamine. ${NO}_{3}^{-}$ $"NO"_3^(-)$ doesn't contribute to the pH because it is the conjugate base of a strong acid (and is thus hardly a base).
$NaClO$ $"NaClO"$ is weakly basic because ${ClO}^{-}$ $"ClO"^(-)$ is the conjugate base of a weak acid, $HClO$ $"HClO"$ (hypochlorous acid). ${Na}^{+}$ $"Na"^(+)$ forms a strong-base hydroxide, and so, it is hardly an acid.
$KCl$ $"KCl"$ is pH-neutral because ${Cl}^{-}$ $"Cl"^(-)$ is the conjugate base of a strong acid (and is thus hardly a base). $K^{+}$ $"K"^(+)$ forms a strong-base hydroxide, and so, it is hardly an acid. Thus, the result is pH-neutral.

The only thing I would question is $C_{2} H_{5} {NH}_{3} CN$ $"C"_2"H"_5"NH"_3"CN"$ . Since $C_{2} H_{5} {NH}_{3}^{+}$ $"C"_2"H"_5"NH"_3^(+)$ and ${CN}^{-}$ $"CN"^(-)$ are a weak acid and a weak base, respectively, I would have to actually check their strengths and make a judgment from there.

The $K_{b}$ $K_b$ of $C_{2} H_{5} {NH}_{2}$ $"C"_2"H"_5"NH"_2$ is about $4.3 \times 10^{- 4}$ $4.3 xx 10^(-4)$ , and the $K_{a}$ $K_a$ of $HCN$ $"HCN"$ is about $6.2 \times 10^{- 10}$ $6.2 xx 10^(-10)$ . At $25^{\circ} C$ $25^@ "C"$ ,

$K_b("CN"^(-)) = K_w/K_a = 10^(-14)/(6.2 xx 10^(-10)) = ul(1.61 xx 10^(-5))$

$K_a("C"_2"H"_5"NH"_3^+) = (10^(-14))/(4.3 xx 10^(-4)) = ul(2.3 xx 10^(-11))$

Since the $K_a$ of $"C"_2"H"_5"NH"_3^(+)$ is much smaller than the $K_b$ of $"CN"^(-)$ , and both ions would be in solution, $"CN"^(-)$ dominates as the base by a factor of about $1000000$ , and this salt is weakly basic.

(It's not strongly basic because the $K_b$ of $"CN"^(-)$ is still small.)

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