Question

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!

Answer 1

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.
[H₃O⁺]=x=7.2335e^-11