Question

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answer 1

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of
`HNO_3`

Formula used :

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the initial molarity and volume of
`C_6H_5NH_2`.

`M_2\text{ and }V_2` are the final molarity and volume of
`HNO_3`.

We are given:

`M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?`

Putting values in above equation, we get:

`0.3403M* 160.0mL=0.0501M* V_2\\\\V_2=1086.79mL`

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of
`C_6H_5NH_2` + Volume of
`HNO_3`

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

`\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M`

Now we have to calculate the pH of the solution.

At equivalence point,

`pOH=(1)/(2)[pK_w+pK_b+\log C]`

`pOH=(1)/(2)[14+4.87+\log (0.0437)]`

`pOH=8.76`

`pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24`

Thus, the pH of the solution is, 5.24