31. Calculate Ks for the a. 5.6x 101 b.40x 10c. 1.0x 10 cyanide ion, CN, (the hy

Question

31. Calculate Ks for the a. 5.6x 101 b.40x 10c. 1.0x 10 cyanide ion, CN, (the hydrolysis constant). (K. for HC-49 x 10 d.2.0x 10 e 2.0 x 10 32. What is the pH of a solution that is 0.10 M in ethanoic acid and 050 M in sodium ethanoate (K, (CH.COOH)- 1.8 x 10]? a 0.70 b. 2.87 c. 4.05d.4.74 e.5.44 33. Which of the following salts will form an acidic aqueous solution? 34. A 25.00-mL sample of propionic acid, CH.CH.COOH, of unknown concentration was titrated with 0.104 M KOH. The equivalence point was reached when 35.31 ml of base had been added. The concentration of the original propionic acid is a. 0.0736 M. b.0128 M. c. 0.147 M. d, 0.162 M. e.0.295 M. 35. What is the solubility product expression for Au(OH)s? 36. The solubility of Poso, is 0.041 g/L. Its molar mass is 303 ghmol. Its K, is a 8.8 x 101. b. 18 x10 c. 1.3 x10. d. 1.7 x 10 e. 4.1 x 102 37. Kap for Fe(IO, is 10-". Mix two solutions, one containing Fe" and the other IO. If, at the instant of mixing. Fe" is 10 M and IO, is 10 M, which one of the following statements is true? a. A precipitate forms because Qw>Kp b. A precipitate forms because QeK d. No precipitate forms because O. Kup. c.No precipitate forms because Q2K e. None of these statements is true. 38. The [OH1 of a saturated solution of Cu(OH) is 8.0 x 10" M. What is the Kap for Cu(OH)? a. 6.4 x 10 b. 1.3x 109 c. 2.6 x 109 d. 5.1 x 10-1 e. 1.3 x 10-18 a. 1.2 x 10 b. 2.9x 10 c. 1.4 x 10 d. 6.9 x 10 e. 1.7 x 10 Ag (ag) +2NHs(ag)Ag(NH)2(ag) 39. The lead ion concentration of a saturated solution of Pblz is 1.2 x 10 M. What is the Ke for lead iodide 40. From a consideration of the reactions K-1.8 x 10 K = 5.6 x 10° Ag (ag)+Cag) AgCI(s) calculate the equilibrium constant for AgCi(s)+2NHs(ag) AgNH)2 (ag)+C(ag) a. 1.0x 10" b. 32x 10 c10x10 d. 1.0x 10 e. 3.1 x10

Explanation / Answer

1.CN- is conjugate base of HCN. Given Ka for HCN= 4.9*10-10, Kb= 10-14/Ka=10-14/(4.9*10-10)=2*10-5 ( d is the correct answer)

2. Henderson-Hasselbach equation is pH= pKa+ log [CH3COO-]/[CH3COOH]

Given [CH3COO-]= 0.5M and [CH3COOH]=0.1M and Ka= 1.8*10-5, pKa= -log Ka= 4.74

pH= 4.74+ log (0.5/0.1)= 5.44 ( E is correct)

3.The following rules need to be remembered.

If the cation comes from a strong base, and the anion comes from a strong acid, the salt will be neutral. From the data given example is NaCl from strong acid HCl and strong base NaOH. Also CaCl2 from strong base Ca(OH)2 and strong acid HCl.
2. If the cation comes from a strong base, and the anion comes from a weak acid, the salt will be basic. NaCN comes from strong base NaOH and weak acid HCN. NaNO2 come from strong base NaOH and weak acid HNO2.
3. If the cation comes from a weak base and the anion comes from a strong acid, the salt will be acidic. FeCl2 comes from Fe(OH)2 a weak base and strong acid HCl. Hence D is correct.
4. if the cation and ion both come from weak acids and bases, it is impossible to predict

IV. Au(OH)3-------->Au+3+3OH-, KSp = [Au+3[ [OH-]3 ( Bi s correct)

5.

For Fe(IO3)3 --àFe+3+3IO3-

Q= [Fe+3] [IO3-]3= 10-4*(10-4)3= 10-16<KSp, no precipitate forms (D is correct)

6.

Cu(OH)2-->Cu+2+2OH-

Given [OH-]= 8*10-7M, [Cu+2]= [OH-]/2= 4*10-7

KSp =[Cu+2] [OH-]2= 8*10-7*(4*10-7)2=1.3*10-19 ( E is corect)

7. PbI2---->Pb+2+2I-

given [Pb+2] = 1.2*10-3M, [I-]= 2*1.2*10-3= 2.4*10-3, KSp= 1.2*10-3*(2.4*10-3)2= 6.912*10-9 ( d is coorect)

8.

For the reaction Ag+(aq)+2NH3(aq)<-> Ag(NH3)2(aq)+

K= [Ag(NH3)2+/ [Ag+] [NH3]2= 1.8*107   (1)

For the reaction Ag+(aq)+Cl-(aq) ->AgCl

K= 1/[Ag+] [Cl-] = 5.6*109 (2), [AgCl] is a solid and it activity is unity.

Eq.1/Eq.2 gives Ag[NH3)2+ [Cl-]/ [NH3]2= 1.8*107/(5.6*109)=3.2*10-3

The K for the desired reaction. ( b is correct)

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