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Question 1 Calculate the percent dissociation of HF (Ka=3.5×10−4) in: a. 0.050 M HF b. 0.50...

Question 1

Calculate the percent dissociation of HF (Ka=3.5×10−4) in:

a. 0.050 M HF

b. 0.50 M HF

Question 2

What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbCl2 from a solution that is 1.00×10−2 M in the chloride ion, Cl−? Ksp for lead(II) chloride is 1.17×10⁻⁵ .

Express your answer with the appropriate units.

Expert Solution

HF dissociates as

HF+ H2O-------->F- + H3O+

Ka= [F-] [H3O+]/[HF]

1. When 0.05F HF is taken, preparing ICE Table ( x= drop in concentration of HF)

Component Initial concentration change Equilibrium concentration

HF 0.05M -x 0.05-x

F- 0 x x

H3O+ 0 x x

Ka= x²/(0.05-x) = 3.5*10-4, when this equation is solved using solver, x=1.31*10^-2,

pH= -log [H3O+]= 2.88

% dissociation = 100*1.31*10^-2/ 0.05 =26.2%

when 0.5M HF is taken, the ICE Table becomes

Component Initial concentration change Equilibrium concentration

HF 0.5M -x 0.5-x

F- 0 x x

H3O+ 0 x x

Ka= x2/(0.5-x)= 3.5*10^-4, when solved using excel, x =0.013, pH= 1.88

% ionization = 100*0.013/0.5=2.6%

3. PbCl2---------> Pb+2+ 2Cl-

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

1.17*10^-5= [Pb+2] *(0.01)²=

[Pb+2]= 0.117M

queen_honey_blossom answered 1 year ago

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