Question

7) 2 OY- A) X- ) - - - D) X->Z- Y- 7 Which one of the following sales when dissolved in water, produces the solution with the lowest A) CAC B) KCI с ква D) AICI pH? D) 2.29 C) 3.89 ) What is the prior a 0.20 M R esolution that has the stepwise dissociation constants Kal-13 10-4 and ka21010-117 A) 5.57 B) 4.59 D) 12.00 10) - 9) What is the pH of a solution prepared by diluting 125.00 mL of 0.020 M Ba(OH)2 with enough water to produce a total volume of 250.00 mL? A) 1.70 B) 12.30 C) 2.00 D) 1% 25.00 mL of 11) – 10) What is the percent dissociation of glycine if the solution has a pH = 8.60 and a pka -9.602 A) 50% B) 5% C) 9% 11) What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0,010 M CH3CO2Na? Assume that the volume of the solutions are additive and that Ka = 1.8 x 10-5 for CH3CO2H. D) 3.74 C) 2.87 B) 4.75 A) 5.74 D-2

Answer 1

Naz D) x-z-Y- B) 2-Y- XZ- A) X-y-z- 9Y-z- x- 7 Which one of the following salts, when dissolved in water, produces the solution with the lowest pH7 A) CaC 7) D) AICI3 9 RbC B) KCI 8) What is the pH of a 0.20 M H2Se solution that has the stepwise dissociation constants Kal-1.3 10-4 and Ka2-1.0 - 10-117 8) D) 2.29 C) 3.89 B) 4.59 A) 5.57 9) What is the pH of a solution prepared by diluting 125.00 mL of 0.020 M Ba(OH)2 with enough 9) water to produce a total volypo of 250.00 mL? A) 1.70 B) 12.30 C) 2.00 D) 12.00 10) What is the percent dissociation of glycine if the solution has a pH = 8.60 and a pKa - 9,60? D) 1% 10) C) 9% B) 5% A) 50% 11) What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 ml. of 0.010 M CH3CO2Na? Assume that the volume of the solutions are additive and that Ka =1.8 x 10-5 for CH3CO2H. 11) A) 5.74 B) 4.75 C) 2.87 D) 3.74 D-2

Explanation :

7) here KCl and CaCl2 , RbCl are the salts of strong acid and strong base . so pH around 7. AlCl3 is the salt of strong acid and weak base . so pH < 7

8)

H2Se   -------------> HSe-   + H+

0.20                          0            0

0.20 - x                     x            x

Ka1 = x^2 / 0.20 - x

1.3 x 10^-4 = x^2 / 0.20 - x

x = 5.1 x 10^-3

pH = 2.29

9)

concentration of Ba(OH)2 = 125 x 0.02 / 250 = 0.01

[OH-] = 0.02 M

pOH = 1.70

pH = 12.30