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:

$0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL$

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.

$\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=\frac{1}{2}[pK_w+pK_b+\log C]$

$pOH=\frac{1}{2}[14+4.87+\log (0.0437)]$

$pOH=8.76$

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

Thus, the pH of the solution is, 5.24