14) Ka for HCN is 4.9 x 10^-10. What is the pH of a 0.068 M aqueous solution of

Question

14) Ka for HCN is 4.9 x 10^-10. What is the pH of a 0.068 M aqueous solution of sodium cyanide? A) 2.96 8) 7.00 C) 11.07 D) 13.24 E) 0.74 15) What is the molar solubility of barium fluoride (BaF2 ) in water? The solubility-product constant for BaF2 is 1.7 x 10^-6 at 25 degree C. A) 1.8 x 10^-3 B) 7.5 x 10^-3 C) 5.7 x 10^-7 D) 6.5 x 10^-7 E) 1.2 x 10^-2 16) The solubility of lead (II) chloride (PbCl2) is 1.6 x 10^-2 M. What is the Ksp of PbCl2? 16) A) 5.0 x 10^-4 8) 3.1 x 10^-7 C) 4.1 x 10^-6 D) 1.6 x 10^-2 E) 1.6 x 10^-5 17) What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The Ksp of PbCl2 is 1.6 x 10^-5. 17) A) 2.0 x 10^-3 B) 7.1 x 10^-4 C) 1.1 x 10^-4 D) 1.6 x 10^-5 E) 1.8 x 10^-4

Explanation / Answer

14. ANSWER : C . 11.07

We have , Ka for HCN = 4.9* 10-10 , Molarity of Aqua NaCN SOLUTION = 0.068M

        We know that aqua NaCN forms from HCN hydrolysis ,

          CN- + H2O ------------> HCN + OH- , Here X = [HCN] = [OH-]

      Kb = X2 / 0.068

     But , Kb = Kw / Ka = 1*10-4 / 4.9*10-10 = 2* 10-5

2*10-5 = X2/0.068

X =[OH-] = 1.2*10-3

POH =log[OH-] = 2.93

pH OF SOLUTION IS , 14-2.93 = 11.07

15 . ANSWER :B : 7.5 * 10-3

     BaF2 ----------->Ba2+ + 2F-     (here ,[Ba+2]= X , [F-] = 2X

    Ksp = [Ba2+][F-]2

   1.7*10-6= X * (2X)2 = 4X3

MOLAR SOLUBILITY OF BaF2 IS , X = 7.5*10-3

    16 . ANSWER : E :1.6 * 10-5

    PbCl2----------> Pb+2 + 2Cl-

   solubility of PbCl2 = 1.6* 10-2M or 1.6*10-2 moles/lit

i.e [Pb+2] = 1.6*10-2 , [Cl-] = 3.2*10-2

      Ksp = (1.6*10-2)(3.2*10-2)

     Ksp = 1.6 * 10-5

17 . ANSWER : B : 7.1 * 10-4

    PbCl2 ----------> Pb+2 + 2Cl-

Ksp = [Pb+2][Cl-]2

   [Pb+] =Ksp / [Cl-]2

          = 1.6*10-5 / (0.15)2

[Pb+] = 7.1 * 10-4

        X = 1.1*10-4

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