Calculate the concentration of H3O(aq) ions present at equilibrium in a solution that is prepared by mixing 100 mL of .35 M HCN and 200 mL of .15 M NaCN.
Kc= 6.2E-10 HCN(aq) + H2O <---> H3O(aq) + CN(aq) (this is an equation from a previous question used for this question also)
Please so all work! Thanks!
the initial concentrations before mixing are [HCN]= 0.35M (A)
and [CN -] = 0.15 (B)
It should calculate the concentration of these compounds after mixing.
For it is multiplied by the given volume, and divided by the total volume of the solution.
Where Vol Total= 100ml +200ml= 300ml
Set up the equilibrium calculation based on
equilibrium identified by the ecuation HCN(aq) + H2O <---> CN(aq) + H3O(aq)
|HCN (aq)||H2O (l)||CN– (aq)||H3O+ (aq)|
|Inicial [ ]||0,11667||------||0.1||0|
|∆ [ ]||-x||------||+x||+x|
|equil [ ]||0.11667-x||------||0.1+x||x|
write the equilibrium equation:
Solve for x
x1=7.2335e-11 ; x2=-0.1000
the value of the positive root is taken.
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