What is the [H+] in 1L solution contains of 0.1M HCOOH (Ka=1.77E-4) & 0.1M HOCN (Ka=3.3E-4)?

Question

anyone knows?
the answer given is [H+] = 0.00712M
but i really cant get the correct way to get the answer, any superior can guide me? thanks you very much

The answer was given by my lecturer, somehow need to use the approximation of dissociation 0.5-x M ~= 0.5 M...

Answer 1

You need to set up 2 ICE tables.
The equilibrium [H+] must satisfy both equilibrium reactions.
This way you get a system of 2 equations

.. .. HCOOH <=> H+ +HCOO- .. .. HOCN<=> H+ +CNO-
Initial .. 0.1 .. .. .. .. .. .. .. .. .. .. .. .. .. .. 0.1
Dissoc. ..x .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. y
Produce .. .. .. .. .. x .. .. .. x .. .. .. .. .. .. .. .. .. .. .. y .. .. ..y
At Equi. 0.1-x .. .. x+y .. .. x .. .. .. .. .. .0.1-y .. ..x+y .. .. y

KaHCOOH = x*(x+y)/(0.1-x) =1.77*10^-4
KaHOCN = y*(x+y)/(0.1-y)= 3.3*10^-4

Solve the system and you get the only positive solution
x=0.00247
y=0.00451

I used the on-line tool at http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=advanced

to solve the system of equations so there is no numerical error.

so [H+]=x+y= 0.00698 M Are you sure about the numbers you give us (both for the solution and the data)