A 1mM solution of zinc chloride is prepared from Zn(NO3)2 (s) and distilled deio

Question

A 1mM solution of zinc chloride is prepared from Zn(NO3)2 (s) and distilled deionized water. The solution is then accidentally left open to the atmosphere over the weekend. No zinc precipitates were observed in the container, the solution pH was 6.3, and after some calculations it was estimated that total dissolved inorganic carbon (H2CO3*+HCO3+CO32) was 0.5 mM. Perform a speciation of zinc in the solution given the following information and neglecting any other potentially occurring complexes:

T = 298 K
H2CO3* H+ + HCO3+; pK1 = 6.3
HCO3 H+ + CO32, pK2 = 10.3

Reaction: Log Kc:
Zn2+ + HCO3 ZnHCO3+ (aq) 12.4
Zn2+ + CO32 ZnCO3 (aq) 5.3

Calculate the concentration for each Zn species in solution.

Explanation / Answer

given pH=6.3

So using henderson-hasselbach equation,

pH=pka+log[Base]/[acid]

gives 6.3=6.3+log [HCO3-]/[H2CO3]

or,[HCO3-]/[H2CO3]=1.............................(1)

Also ,6.3=10.3++log [CO32-]/[HCO3-]

or,log [CO32-]/[HCO3-]=-4........................(2)

or,[CO32-]/[HCO3-]=10^-4=0.0001

given,[H2CO3] +[HCO3]+ [CO32] = 0.5 mM.....................(3)

from (1) and (3),[H2CO3] +[HCO3]+ [CO32] = 0.5 mM.

2HCO3]+ [CO32] = 0.5 mM.

from (2),

[CO32-]/[HCO3-]=0.0001

or,[CO32-]=0.0001*[HCO3-]

or. 2[HCO3]+ 0.0001*[HCO3-] = 0.5 mM.

or,2.0001[HCO3]=0.5mM

or,[HCO3]=0.249M

and [CO32-]=0.0001*0.249=0.249*10^-4M

calculation of [ZnHCO3+]

kc=[ ZnHCO3+]/[Zn2+] [HCO3-]

log kc=12.4

kc=10^12.4=2.511*10^12

As kc >>> so lets assume the rxn goes to completion ,

ICE                      [Zn2+] [HCO3]                    [ZnHCO3+]

initial                    1mM                                          0.249M                        0

change                 1mM-0.249mM=0.751              0.249-0.249=0             0.249

equilibrium                 0.751+x                                       x                          0.249-x

keq=1/kc=[Zn2+] [HCO3-]/[ ZnHCO3+]

1/2.511*10^12=( 0.751+x )x /(0.249-x)

0.4*10^-12=( 0.751+x )x /(0.249-x)

or,0.4*10^-12=( 0.751 )x /(0.249) [neglecting x <<<]

or, x=0.132*10^-12M

[ZnHCO3+]=0.249-x=0.249-0.132*10^-12=0.249M

calcultion of [ZnCO3]

log kc=5.3

kc=1.99*10^5

                       [Zn2+]                                          [CO32] [ ZnCO3 ]

initial               1mM                      0.249*10^-4M               0

completion       1mM-0.249*10^-4M   0                           0.249*10^-4M

change                   +x                                                              +x                                   -x

euilibrium                  (1-0.249*10^-4M)                                         x                              ( 0.249*10^-4M-x)

                             =1mM(approx)

so kc=1.99*10^5=( 0.249*10^-4M-x)/1mM*x

or,1.99*10^5=( 0.249*10^-4M-x)/1mM*x

or, 1.99*10^5=( 0.249*10^-4M)/1mM*x

or,x=0.125*10^-9M

so,[ZnCO3]= ( 0.249*10^-4M-x)= ( 0.249*10^-4M-0.125*10^-9M=0.249*10^-4M

,[ZnCO3]=0.249*10^-4M

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