A neutralization reaction is when an acid and a base reaction to kind water and a salt and involves the mix of H+ ions and also OH- ions to generate water. The neutralization of a strong acid and strong base has actually a pH same to 7. The neutralization that a strong acid and also weak basic will have a pH of less than 7, and also conversely, the resulting pH once a solid base neutralizes a weak acid will be greater than 7.

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When a systems is neutralized, it way that salts are created from equal weights of acid and base. The amount of acid required is the amount that would give one mole of protons (H+) and the lot of base required is the amount that would give one mole of (OH-). Since salts are formed from neutralization reactions with tantamount concentrations of weights of acids and bases: N parts of acid will always neutralize N parts that base.

Table (PageIndex1): The most common strong acids and bases. Most every little thing else not in this table is taken into consideration to be weak. Strong AcidsStrong Bases

Strong Acid-Strong base Neutralization

Consider the reaction in between (ceHCl) and also (ceNaOH) in water:

This deserve to be created in regards to the ion (and canceled accordingly)

When the spectator ions space removed, the net ionic equation reflects the (H^+) and also (OH^-) ions creating water in a strong acid, solid base reaction:

(H^+_(aq) + OH^-_(aq) leftrightharpoons H_2O_(l) )

When a strong acid and a strong base completely neutralize, the pH is neutral. Neutral pH way that the pH is same to 7.00 in ~ 25 ºC. In ~ this suggest of neutralization, there space equal amounts of (OH^-) and (H_3O^+). There is no excess (NaOH). The solution is (NaCl) at the equivalence point. As soon as a strong acid fully neutralizes a strong base, the pH that the salt equipment will constantly be 7.

Weak Acid-Weak basic Neutralization

A weak acid, weak basic reaction deserve to be displayed by the network ionic equation example:

(H^+ _(aq) + NH_3(aq) leftrightharpoons NH^+_4 (aq) )

The equivalence allude of a neutralization reaction is once both the acid and the base in the reaction have actually been totally consumed and neither that them room in excess. Once a strong acid neutralizes a weak base, the result solution"s pH will certainly be less than 7. As soon as a solid base neutralizes a weak acid, the result solution"s pH will certainly be higher than 7.

Table 1: pH levels at the Equivalence allude Strength that Acid and BasepH Level
Strong Acid-Strong Base 7
Strong Acid-Weak Base 7
Weak Acid-Weak Base pH K_b) pH =7 if (K_a = K_b) pH >7 if (K_a


One of the most common and widely used means to complete a neutralization reaction is with titration. In a titration, an mountain or a basic is in a flask or a beaker. We will present two instances of a titration. The first will be the titration that an mountain by a base. The 2nd will it is in the titration the a basic by one acid.

Example (PageIndex1): Titrating a Weak Acid

Suppose 13.00 mL that a weak acid, v a molarity the 0.1 M, is titrated v 0.1 M NaOH. How would we draw this titration curve?


Step 1: First, we require to discover out whereby our titration curve begins. To execute this, we uncover the early stage pH the the weak acid in the manufacturer before any kind of NaOH is added. This is the allude where ours titration curve will certainly start. To find the initial pH, we very first need the concentration that H3O+.

Set increase an ice cream table to find the concentration that H3O+:

(HX) (H_2O) (H_3O^+) (X^-)
Initial 0.1M
Change -xM +xM +xM
Equilibrium (0.1-x)M +xM +xM


Solve for pH:


Step 2: To accurately attract our titration curve, we need to calculate a data suggest between the beginning point and the equivalence point. To execute this, we settle for the pH once neutralization is 50% complete.

Solve for the mole of OH- the is added to the beaker. We can to do by very first finding the volume of OH- included to the mountain at half-neutralization. 50% that 13 mL= 6.5mL

Use the volume and also molarity to deal with for moles (6.5 mL)(0.1M)= 0.65 mmol OH-

Now, deal with for the mole of mountain to it is in neutralized (10 mL)(0.1M)= 1 mmol HX

Set increase an ice table to determine the equilibrium concentration of HX and also X:

(HX) (H_2O) (H_3O^+) (X^-)
Initial 1 mmol
Added Base 0.65 mmol
Change -0.65 mmol -0.65 mmol -0.65 mmol
Equilibrium 0.65 mmol 0.65 mmol

To calculation the pH at 50% neutralization, usage the Henderson-Hasselbalch approximation.


pH=pKa+ log<0.65mmol/0.65mmol>


Therefore, as soon as the weak mountain is 50% neutralized, pH=pKa

Step 3: Solve for the pH in ~ the equivalence point.

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The concentration the the weak acid is half of its initial concentration once neutralization is complete 0.1M/2=.05M HX

Set up an ice cream table to recognize the concentration of OH-:

Initial 0.05 M
Change -x M +x M +x M
Equilibrium 0.05-x M +x M +x M


Since Kw=(Ka)(Kb), we can substitute Kw/Ka in place of Kb to acquire Kw/Ka=(x^2)/(.05)


Step 4: Solve because that the pH ~ a bit an ext NaOH is included past the equivalence point. This will offer us specific idea of wherein the pH levels off at the endpoint. The equivalence suggest is when 13 mL the NaOH is added to the weak acid. Let"s uncover the pH after 14 mL is added.

Solve because that the moles of OH-

< (14 mL)(0.1M)=1.4; mmol OH^->

Solve for the mole of acid

<(10; mL)(0.1;M)= 1;mmol ;HX>

collection up an ice table to recognize the (OH^-) concentration:

(HX) (H_2O) (H_3O^+) (X^-)
Initial 1 mmol
Added Base 1.4 mmol
Change -1 mmol -1 mmol 1 mmol
Equilibrium 0 mmol 0.4 mmol 1 mmol