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 ?

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

#NH_4Cl#: #color(red)("Acidic solution")#.
#NH_4Cl(aq)->NH_4^(+)(aq)+Cl^(-)(aq)#

#NH_4CH_3COO#: #color(green)("Neutral solution")#.
#(K_b)_(NH_3)=1.8xx10^-5# and #(K_a)_(CH_3COOH)=1.8xx10^-5# therefore, #K_a=K_b#
#NH_4CH_3COO(aq)->NH_4^(+)(aq)+CH_3COO^(-)(aq)#

#KCl#: #color(green)("Neutral solution")#
#KCl(aq)->K^(+)(aq)+Cl^(-)(aq)#

#KCH_3COO#: #color(blue)("Basic solution")#.
#KCH_3COO(aq)->K^(+)(aq)+CH_3COO^(-)(aq)#

#NH_4NO_2#: #color(red)("Acidic solution")#.
#(K_b)_(NH_3)=1.8xx10^-5# and #(K_a)_(HNO_2)=4.5xx10^-4# therefore, #K_a>K_b#
#NH_4NO_2(aq)->NH_4^(+)(aq)+NO_2^(-)(aq)#

#KNO_2#: #color(blue)("Basic solution")#.
#KNO_2(aq)->K^(+)(aq)+NO_2^(-)(aq)#

#"Pyridinium"Cl#: #color(red)("Acidic solution")#.
#PyrNHCl(aq)->PyrNH^(+)(aq)+Cl^(-)(aq)#

#"Pyridinium"CH_3COO#: #color(red)("Acidic solution")#.
#(K_b)_("pyridine")=1.7xx10^-9# and #(K_a)_(CH_3COOH)=1.8xx10^-5# therefore, #K_a>K_b#
#PyrNHCH_3COO(aq)->PyrNH^(+)(aq)+CH_3COO^(-)(aq)#

#"Pyridinium"NO_2#: #color(red)("Acidic solution")#.
#(K_b)_("pyridine")=1.7xx10^-9# and #(K_a)_(HNO_2)=4.5xx10^-4# therefore, #K_a>K_b#
#PyrNHNO_2(aq)->PyrNH^(+)(aq)+NO_2^(-)(aq)#

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.