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

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When a equipment is neutralized, it means that salt are created from same weights the acid and base. The lot of acid essential is the amount the would offer one mole of proton (H+) and also the lot of base required is the amount the would provide one mole that (OH-). Due to the fact that salts are created from neutralization reactions with indistinguishable concentrations of weights the acids and bases: N parts of acid will constantly neutralize N parts of base.

Table (PageIndex1): The most common strong acids and also bases. Most everything else no in this table is considered to be weak. Strong AcidsStrong Bases

Strong Acid-Strong basic Neutralization

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

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

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

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

When a strong acid and also a strong base totally neutralize, the pH is neutral. Neutral pH way that the pH is equal to 7.00 at 25 ºC. At this allude of neutralization, there are equal amounts of (OH^-) and also (H_3O^+). There is no overabundance (NaOH). The systems is (NaCl) in ~ the equivalence point. As soon as a strong acid totally neutralizes a strong base, the pH that the salt systems will constantly be 7.

Weak Acid-Weak basic Neutralization

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

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

The equivalence suggest of a neutralization reaction is when both the acid and the base in the reaction have actually been fully consumed and neither of them space in excess. When a solid acid neutralizes a weak base, the result solution"s pH will be much less than 7. When a solid base neutralizes a weak acid, the resulting solution"s pH will certainly be higher than 7.

Table 1: pH levels at the Equivalence allude Strength of Acid and also 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 many common and widely used means to complete a neutralization reaction is with titration. In a titration, an mountain or a base is in a flask or a beaker. We will present two examples of a titration. The first will be the titration the an mountain by a base. The 2nd will it is in the titration that a base by an acid.

Example (PageIndex1): Titrating a Weak Acid

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


Step 1: First, we require to uncover out where our titration curve begins. To execute this, we find the early stage pH that the weak mountain in the beaker before any NaOH is added. This is the suggest where ours titration curve will certainly start. To discover the early pH, we an initial need the concentration that H3O+.

Set increase an ice table to find the concentration of 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 draw our titration curve, we have to calculate a data suggest between the beginning point and the equivalence point. To carry out this, we solve for the pH as soon as neutralization is 50% complete.

Solve for the mole of OH- that is added to the beaker. We deserve to to carry out by first finding the volume that OH- included to the acid at half-neutralization. 50% that 13 mL= 6.5mL

Use the volume and also molarity to solve for mole (6.5 mL)(0.1M)= 0.65 mmol OH-

Now, fix for the mole of acid to be neutralized (10 mL)(0.1M)= 1 mmol HX

Set up an ice table to identify the equilibrium concentrations of HX and 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 calculate the pH in ~ 50% neutralization, use 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 because that the pH at the equivalence point.

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The concentration that the weak mountain is half of its initial concentration when neutralization is finish 0.1M/2=.05M HX

Set up an ice 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 have the right to substitute Kw/Ka in ar of Kb to get Kw/Ka=(x^2)/(.05)


Step 4: Solve for the pH after a bit much more NaOH is included past the equivalence point. This will provide us an accurate idea of where the pH levels off at the endpoint. The equivalence suggest is once 13 mL that NaOH is included to the weak acid. Let"s discover the pH after 14 mL is added.

Solve because that the mole of OH-

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

Solve for the moles of acid

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

set up an ice cream table to identify 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