Question

63-74) A 40.0 mL sample of 0.18 M HCl is titrated with 0.36 M C&HgNH2. Determine the pH at these points: At the beginning (before base is added) show work a. kbr 3.9x10 pH After adding 15.00 mL of KOH b. show work pH After adding 30.00 mL of KOH c. show work nH

Answer 1

a) Here, the solution only contains 0.18M HCl

HCl will dissociate completely to give 0.18mol H⁺ in the solution.

By definition, pH is the negative log of hydrogen ion concentration. Therefore, in this case

pH = -log(0.18) = 0.7447

b) Let us assume that base added is 0.36M C₆H₅NH₂ and not KOH (details of KOH solution are not given)

First let us find our the moles of HCl remaining in the solution

Molarity of C₆H₅NH₂ = 0.36  

Volume of C₆H₅NH₂ added = 15ml

Therefore, number of moles of C₆H₅NH₂ added = 0.36 X 0.015 = 0.0054

Therefore, number of moles of HCl consumed =0.0054

Initial moles of HCl present = 0.18 X 0.40 = 0.0072

Therefore, moles of HCl present = 0.0072 - 0.0054 = 0.0018

Now, volume of solution is 40ml (initial) + 15ml (added base) = 55ml

Therefore, molarity of HCl = 0.0018/0.055 = 0.0327M

Therefore, pH of the solution = -log(0.0327) = 1.485

c) In this case, C₆H₅NH₂ added is 0.36 X 0.03 = 0.0108 moles, which is greater than the number of moles of HCl present. This means that the reaction has passed it's equivalence point and now has excess of C₆H₅NH₂. Therefore, we should calculate the excess of C₆H₅NH₂ and it's conjugate acid C₆H₅NH3⁺ to determine the pH

Here, Number of moles of C₆H₅NH₂ converted to C₆H₅NH3⁺ = 0.0072 (equal to the number of moles of HCl added)

Therefore, number of moles of C₆H₅NH3⁺ = 0.0072

Number of moles of unconverted C₆H₅NH₂ = 0.0108 - 0.0072 = 0.0036

Volume of the solution is now 40ml (intial) + 30ml (added base) = 70ml

Therefore, molarity of C₆H₅NH3⁺ = 0.0072/0.070 = 0.102M

Molarity of unconverted C₆H₅NH₂ = 0.0036/0.070 = 0.0514

Now, using Henderson- Hasselbalch approximation

+ CH5NH3 pOH pKblog CHNH

Putting above values into this equation

0.102 pOH log(3.9X1010)log0514

Therefore, pOH = 9.710

pH = 14- pOH

Therefore pH = 14 - 9.710 = 4.29