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∙ 13y agoAcetic acid (Ethanoic acid) is a weak acid, and when reacted with a strong base like Potassium hydroxide, it establishes an equilibrium:
CH3COOH + KOH <=> CH3COOK + H2O
The reaction mixture contains all four products in different proportions, and as such, an acid buffer is created. When an acid is added, the CH3COO- ions (those mixed with the K+ ions) 'mop up' the H+ ions from the acid. When a base is added, the H+ ions from the CH3COOH 'mop up' the OH- ions so the pH is little affected.
NB. pH=-log10(H+)
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∙ 12y agoWiki User
∙ 13y agoCH3COOH (aq) + OH- -----> CH3COO- (aq) + H2O
The potassium is a spectator ion and can thus be deleted from the equation.
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∙ 9y agoPotassium oxalate has the chemical formula C2K2O4. When reacting with acetic acid (CH3COOH), the reaction is K2C2O4 + CH3COOH = 2 KCH3COO + HC2O4.
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∙ 11y agoCr^3+ + 3K2C2O4^2- >>> [Cr(C2O4)3]^3-
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∙ 10y agoThese reagents undergo a neutralization reaction. The products are potassium acetate (salt) and water.
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∙ 14y agoany strong acid..
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∙ 6y agoCH3COOH + KOH ==> CH3COOK + H2O
The preparation equation depends on the route by which this compound is prepared. A simple route is neutralization of acetic acid with potassium hydroxide: KOH + CH3COOH --> H2O + K+CH3COO-
HC2H3O2(aq) + KOH(aq) → KC2H3O2(aq) + H2O(l)...... I don't know about Net Ionic, but i think this is correct
Water, potassium and acetate ions. If you're adding equal amounts of both, the final solution will have a pH greater than 7.
CH3COOH + KOH = CH3COOK + H2O
LiC2H3O2 + H2O
The preparation equation depends on the route by which this compound is prepared. A simple route is neutralization of acetic acid with potassium hydroxide: KOH + CH3COOH --> H2O + K+CH3COO-
HC2H3O2(aq) + KOH(aq) → KC2H3O2(aq) + H2O(l)...... I don't know about Net Ionic, but i think this is correct
Water, potassium and acetate ions. If you're adding equal amounts of both, the final solution will have a pH greater than 7.
CH3COOH + KOH = CH3COOK + H2O
LiC2H3O2 + H2O
C_2_H_4_O_2_ (aq) + OH^-^ (aq) --> C_2_H_3_O_2_^-^ (aq) + H_2_O (l)
Potassium hydroxide and acetic acid.
CH3COOH + OH ---> CH3COO + H2O CH3COOH stays as a molecule because it is a weak acid
Acetic acid is an organic acid an d it reacts with potassium hydroxide to form salt and water Its neuatralisation reaction CH3CooH + KOH -------> CH3COOK + H2O
CH3COOH(aq) + KOH(aq) --->CH3COOK(aq) + H2O(l)You can obtain the CH3COOK by removal of the water.
KOH(aq)+HC2H3O2(aq)---- H2O(l)+KC2H3O2(aq)
CH3COOH+NaOH=CH3COONa+H2O