Neutralization Reaction - Definition, Equations and Applications
Neutralization reactions are the reaction between acid and base. The products formed are water and salt. It is called so because the acid and base neutralize each other to form water and salt. The H+ and OH- react to form H2O.
The strength and the volume of the acid and base are required to carry out a neutralization reaction. If the strengths are known, then we can find the volumes are which they will completely neutralize each other and vice versa. For this, we can use the formula,
When a strong acid neutralizes a strong base, the resultant is neutral because all the H+ ions react with all the OH-ions to form water. There are no surplus H+ or OH- ions present in the resultant solution. Its pH is 7.
Strong acid: HA +H2O → A- (aq) + H3O+ (aq) Strong base: BOH + H2O→ B+ (aq) + OH-(aq)
2. STRONG ACID AND WEAK BASE
Weak bases don’t dissociate only to about 5-10%. Here the amount of OH- ions is very less compared to the amount of H+ ions available. Therefore, the resultant solution is acidic. Its pH lies between 3-6.
Weak base: BOH + H2O ↔ B+ (aq) + OH-(aq)
(Or) B + H2O ↔ BH+ (aq) + OH-(aq) Examples of weak bases include Mg (OH) 2, Ca (OH) 2, and Ba (OH) 2.
Kb = [OH-] [B+] / [BOH]
Taking log on both sides you get:
Log Kb = log([OH-] [B+] / [BOH]) Splitting the log terms: Log Kb = log [OH-] + log ([B+] / [BOH]) Since pKb = -log10 Kb and pOH = -log10 [OH-], therefore: -pKb = – pOH + log ([B+]/ [BOH]) POH = pKb + log ([B+]/ [BOH]) This equation can also be written as: pOH = pKb + log [(salt)/(base)]
3. STRONG BASE AND WEAK ACID
Weak acids don’t dissociate entirely in aqueous conditions. The OH- ions concentration is much higher compared to the free H+ ions. Therefore, the resultant solution is basic with a pH of around 8-11.
AH +H2O ↔ A-(aq) + H3O+ (aq)
Example of weak acids include; acetic acid and all the other organic acids.
Following the law of dissociation, the acid dissociation constant Ka can be defined according to the equation as,
Ka = [H+] [A–]/ [HA] Taking log on both sides you get: Log Ka = log ([H+] [A–]/ [HA]) Splitting the log terms; Log Ka = log [H+] + log ([A–]/ [HA]) Since pKa = -log10 Ka and pH = -log10 [H+], therefore: -pKa = – pH + log ([A–]/ [HA] pH = pKa + log ([A–]/[HA]) This equation can also be written as: pH = pKa + log [(salt)/(acid)]
4. WEAK ACID AND WEAK BASE
Here since both, acid and base, is weak neither of them dissociates completely and so neutralization does not occur. Equilibrium is the state when the rate of formation of product is equal to the rate of formation of the reactants.
1. It is used in fertilization as the crop cannot grow in acidic soil. 2. On bee sting, using baking soda to neutralize the effect of formic acid that it has released in our body. 3. When we suffer from acidity, we take antacid tablet Mg (OH) 2 to neutralize the effect of HCl produced in our stomach.