Question

14 of 18 Questions Assignment Score: 60.2% Resources Hint Check Answer 100% Question 14 of 18 > 10 Question 1ofAllempls Correct The pki, values for the dibasic base B are pKb = 2.10 and pK 12 = 7.66. 046 11 Question O of o Attempts Calculate the pH at each of the points in the titration of 50.0 mL of a 0.70 M B(aq) solution with 0.70 M HCl(aq). What is the pH before addition of any HCI? 95% 12 Question 2 of Attempts Correct 0% What is the pH after addition of 25.0 mL HCI? 13 Question O of Attempts 0% What is the pll after addition of 50.0 mL HCI? 14 Question o of Attempts 959 What is the pH after addition of 75.0 mL HCI? 15 Question 2 of Attempts Correct 100% 16 Question 1 of Attempts What is the pH after addition of 100.0 mL HCI? Correct 100% 17 Question 1 ofco Attempts Correct 100% 18 Question 1 of Attempts Correct

Answer 1

Answer:

The given concentrations are 0.7M diprotic base and 0.7 M HCl

since the concentrations are equal, the first equivalence point for 50 mL of the base = 50 mL of HCl

                                                   the second equivalence point = 100 mL of HCl

therefore the first half equivalence point is 25 mL

the second half equivalence point is 75 mL

we know that the first half equivalence point pH = 14 - pKb1 = 14 - 2.1 - 11.9

                   the second half equivalence point pH = 14 -pKb2 = 14 - 7.66 = 6.34

                  the pH at first equivalence point = 14 - (pKb1+pKb2)/2 = 9.12

The initial pH is givenby the dissociation of the base according to the following equations

B + H₂O $\rightleftharpoons$ BH⁺+ OH^- pKb1 = 2.1 therefore kb1 = 0.00795

0.7 -x           x         x

x²/0.7-x = 0.00795 solving the quadratic equation we get x = 0.0705

therefore the concentration of OH- ion = 0.0705

BH⁺ + H₂O $\rightleftharpoons$ BH²⁺ + OH^-   pKb2 = 7.66 therefore Kb2 = 2.19 x 10^-8

0.0705 -x          x          x

x²/0.0705-x     = 2.19 x 10^-8

solving by neglecting x in the denominator we get x = 0.000039

Hence the total OH- concentration = 0.070539 and hence pOH = 1.15

therefore pH -= 14 - 1.15 = 12.85

The pH after addition of 100 mL HCl is the second equivalence point

Here all the Base will be converted as BH²⁺ and its concentration will be 0.7 x 50 / 150 = 0.233 M since the total volume would be 150 mL

consider the equation

BH²⁺ + H₂O $\rightleftharpoons$ BH⁺ + H₃ O+

the pKa value of this equation is given by 14 - pKb2 = 14 - 7.66 = 6.34

and hence the Ka value = 4.57 x 10^-7

now calculating the H+ ion concentration

x² / 0.233 -x = 4.57 x 10^-7 solving for x neglecting x in the denominator

we get x = 3.26 x 10^-4

H+ concentration = 3.26 x 10^-4 which implies the pH = 3.49

Thus

PH initial = 12.85

After adding 25 ml HCl = 11.9

After adding 50 ml HVl = 9.12

After adding 75 ml HCl = 6.43

After adding 100 ml HCl = 3.49