Answer : The pH of the solution is, 5.24
Explanation :
First we have to calculate the volume of [tex]HNO_3[/tex]
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of [tex]C_6H_5NH_2[/tex].
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of [tex]HNO_3[/tex].
We are given:
[tex]M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL[/tex]
Now we have to calculate the total volume of solution.
Total volume of solution = Volume of [tex]C_6H_5NH_2[/tex] + Volume of [tex]HNO_3[/tex]
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.
[tex]\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M[/tex]
Now we have to calculate the pH of the solution.
At equivalence point,
[tex]pOH=\frac{1}{2}[pK_w+pK_b+\log C][/tex]
[tex]pOH=\frac{1}{2}[14+4.87+\log (0.0437)][/tex]
[tex]pOH=8.76[/tex]
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24[/tex]
Thus, the pH of the solution is, 5.24
