# HCl ,CH3COOH, HNO2,NH3,KOH, pyridine. Write as many salts formula. For each salt,write chemical equation when salt dissolved in water. Is the solution acidic neutral or basic ?

Jan 26, 2016

See explanation.

#### Explanation:

The salts that could form and their corresponding aqueous solutions are:

$N {H}_{4} C l$: $\textcolor{red}{\text{Acidic solution}}$.
$N {H}_{4} C l \left(a q\right) \to N {H}_{4}^{+} \left(a q\right) + C {l}^{-} \left(a q\right)$

$N {H}_{4} C {H}_{3} C O O$: $\textcolor{g r e e n}{\text{Neutral solution}}$.
${\left({K}_{b}\right)}_{N {H}_{3}} = 1.8 \times {10}^{-} 5$ and ${\left({K}_{a}\right)}_{C {H}_{3} C O O H} = 1.8 \times {10}^{-} 5$ therefore, ${K}_{a} = {K}_{b}$
$N {H}_{4} C {H}_{3} C O O \left(a q\right) \to N {H}_{4}^{+} \left(a q\right) + C {H}_{3} C O {O}^{-} \left(a q\right)$

$K C l$: $\textcolor{g r e e n}{\text{Neutral solution}}$
$K C l \left(a q\right) \to {K}^{+} \left(a q\right) + C {l}^{-} \left(a q\right)$

$K C {H}_{3} C O O$: $\textcolor{b l u e}{\text{Basic solution}}$.
$K C {H}_{3} C O O \left(a q\right) \to {K}^{+} \left(a q\right) + C {H}_{3} C O {O}^{-} \left(a q\right)$

$N {H}_{4} N {O}_{2}$: $\textcolor{red}{\text{Acidic solution}}$.
${\left({K}_{b}\right)}_{N {H}_{3}} = 1.8 \times {10}^{-} 5$ and ${\left({K}_{a}\right)}_{H N {O}_{2}} = 4.5 \times {10}^{-} 4$ therefore, ${K}_{a} > {K}_{b}$
$N {H}_{4} N {O}_{2} \left(a q\right) \to N {H}_{4}^{+} \left(a q\right) + N {O}_{2}^{-} \left(a q\right)$

$K N {O}_{2}$: $\textcolor{b l u e}{\text{Basic solution}}$.
$K N {O}_{2} \left(a q\right) \to {K}^{+} \left(a q\right) + N {O}_{2}^{-} \left(a q\right)$

$\text{Pyridinium} C l$: $\textcolor{red}{\text{Acidic solution}}$.
$P y r N H C l \left(a q\right) \to P y r N {H}^{+} \left(a q\right) + C {l}^{-} \left(a q\right)$

$\text{Pyridinium} C {H}_{3} C O O$: $\textcolor{red}{\text{Acidic solution}}$.
${\left({K}_{b}\right)}_{\text{pyridine}} = 1.7 \times {10}^{-} 9$ and ${\left({K}_{a}\right)}_{C {H}_{3} C O O H} = 1.8 \times {10}^{-} 5$ therefore, ${K}_{a} > {K}_{b}$
$P y r N H C {H}_{3} C O O \left(a q\right) \to P y r N {H}^{+} \left(a q\right) + C {H}_{3} C O {O}^{-} \left(a q\right)$

$\text{Pyridinium} N {O}_{2}$: $\textcolor{red}{\text{Acidic solution}}$.
${\left({K}_{b}\right)}_{\text{pyridine}} = 1.7 \times {10}^{-} 9$ and ${\left({K}_{a}\right)}_{H N {O}_{2}} = 4.5 \times {10}^{-} 4$ therefore, ${K}_{a} > {K}_{b}$
$P y r N H N {O}_{2} \left(a q\right) \to P y r N {H}^{+} \left(a q\right) + N {O}_{2}^{-} \left(a q\right)$

Here is a video that explains how to determine the type of the solution based on its composition:

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Acids & Bases | Acid - Base Properties of Salts.