5. Find the pH of the following solutions generated from the materials in the bo

Question

5. Find the pH of the following solutions generated from the materials in the bottles shown below: 200.0 mL of 0.0750 M HCI Bottle A Bottle B 200.0 mL of 0.200 M hydrazine (H2N-NH2, pKb 5.88) Bottle C 100.0 g of solid hydrazine hydrochloride. H N-NH3C1 a. Solution #1: 6.80 g of the solid in Bottle C dissolved in water to give a total volume of 500.0 mL b. Solution #2: Mix 75.0 mL of Solution #1 with 75.0 mL of the solution in Bottle B. c. Solution #3: Mix 15.0 mL of the solution in Bottle A with 50.0 mL of Solution #2.

Explanation / Answer

a)Solution 1

It is a salt of weak base with strong acid and its solution pH is given by

pH = 1/2 pkw -1/2pkb -1/2 log C

So we calculate the concentration c of the solution

concentration = (weight/molar mass ) x (1000/V(mL))

= (6.8/69.5) (1000/500)

= 0.1956 M

The pH 0f the solution = 7 -(5.88/2) -1/2 log (0.1956)

=3.70

b) We mixed 75ml of 0.1956M salt solution with 75ml of 0.2 M base solution of pkb =5.88

The pH of this buffer is given by Hendersen equation

pH = 14 -{pkb + log [conjugate acid]/[base]}

= 14 - { 5.88 + log [75x0.1956]/[75x0.2]

= 8.1296

c) NOw the buffer solution 50 mL is mixed with 15 ml of 0.075M Hcl

B + HCl ------> BH+ + Cl-

0.2 x50 0 0.2 x50 0 before

15x 0.075 change

8.875 0 11.125 - after addition of HCl

8.875/65 11.125/65 concentrations aftr

Thus pH = 14 - { 5.88 + log ]11.125/8.825]

   =8.021

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