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Question: Calculate the pH of a solution prepared by mixing 100mL of 0.4M \[HCl\] with 100 mL of 0.4 M \[N{H_3...

Calculate the pH of a solution prepared by mixing 100mL of 0.4M HClHCl with 100 mL of 0.4 M NH3N{H_3}. Hydrolysis constant of ammonium chloride is 5.6×10105.6 \times {10^{ - 10}}

Explanation

Solution

In order to find the pH of a solution which is prepared by mixing HClHCl and NH3N{H_3}, we must first know what a pH is. pH is a type of scale which is used to measure the acidity or basicity of an aqueous solution. The expanded form for pH is the power of Hydrogen. pH will indicate the concentration of the H+{H^ + } ions.

Complete Solution :
Let us first understand about pH. pH is a type of scale which is used to measure the acidity or basicity of an aqueous solution. The expanded form for pH is the power of Hydrogen.
Different types of substance will have different pH.
- pH for an acidic substance is below 7.
- pH for a basic substance will be above 7.
- pH for a neutral substance will be equal to 7.

Now let us move onto the problem given.
The aqueous solution of the ammonia is the ammonium hydroxide.
The equilibrium reaction between HClHCland NH4OHN{H_4}OH is given below:
HCl+NH4OHNH4Cl+H2OHCl + N{H_4}OH \rightleftharpoons N{H_4}Cl + {H_2}O

We have to remember that as were mixing the strong acid, i.e. HClHCl with a weak base, i.e. NH3N{H_3}, the pH of the resulting solution will never be equal to 7. From the question we can see that equal volumes of NH3N{H_3} and HClHCl are reacting, therefore the number of moles of the reactants undergoing neutralisation will also be equal.
Total volume of the liquids, i.e. NH3N{H_3} and HClHCl = 100 + 100 = 200mL.
Therefore, the number of moles of NH4ClN{H_4}Cl = 40200=0.2M\dfrac{{40}}{{200}} = 0.2M
The equilibrium reaction between NH4+NH_4^ + and H2O{H_2}O is given as
NH4++H2OH++NH4OHNH_4^ + + {H_2}O \rightleftharpoons {H^ + } + N{H_4}OH

Now we have to look for the base dissociation constant kb{k_b} of NH3N{H_3} at the room temperature.
kb=1.78×105{k_b} = 1.78 \times {10^{ - 5}}
pH=70.5(pKb+logC)pH = 7 - 0.5(p{K_b} + \log C)
pH=70.5[5log(1.78)+log(0.2)]pH = 7 - 0.5[5 - \log (1.78) + \log (0.2)]
pH=4.96pH = 4.96
The pH of the solution prepared by mixing NH3N{H_3} and HClHCl is 4.96.

Note: We must remember that pH and pOH are different from one another. We can say that they are opposite of each other. pOH indicates the concentration of the OHO{H^ - } ions.
- pOH of the acidic substance is above 7, while pH will be below 7.
- pOH of the basic substance is below 7, while pH will be above 7.