Question

Calculate the pH of the solution resulting from the addition of 85.0 mL of 0.35 M HCl to 30.0 mL of 0.40 M aniline (C₆H₅NH₂). K_b (C₆H₅NH₂) = 3.8 x 10^-10

9.09

4.64

4.19

0.81

1.75 Please fully explain your answer

Answer 1

Solution :-

Given data

Volume of HCl = 85.0 ml

Molarity of HCl = 0.35 M

Volume of C6H5NH2 = 30.0 ml

Molarity of the C6H5NH2 = 0.40 M          

pH after mixing of solution

After the mixing the solutions following reaction will takes place

HCl + C6H5NH2 ----------- > C6H5NH3^+(aq) + Cl^-(aq)

Now lets calculate moles of the HCl and C6H5NH2 using their molarities and volumes

Moles = molarity * volume in liter

Moles of HCl = 0.35 mol per L * 0.085 L =0.02975 mol

Moles of C6H5NH2 = 0.40 mol per L * 0.030 L = 0.012 mol C6H5NH2

Mole ratio of the HCl to C6H5NH2 is 1 : 1

Here moles of the C6H5NH2 are less than moles of HCl

Therefore HCl is the excess reagent and C6H5NH2 is the limiting reagent

Therefore after reaction some of the HCl will remain in the solution

So lets calculate the moles of the HCl remained after the reaction

Moles of HCl remained after reaction = 0.02975 mol-0.012 mol = 0.01775 mol

Now lets calculate new molarity of the HCl at total volume

Total volume = 85.0 ml + 30.0 ml = 115.0 ml = 0.115 L

Molarity = moles / volume in liter

New molarity of the HCl = 0.01775 mol / 0.115 L = 0.15434 M

Since HCl is the strong acid therefore it dissociate 100 %

Therefore concentration of the H+ = 0.15434 M

Now lets calculate the pH

pH= - log[H+]

pH= -log [0.15434]

pH= 0.81