here is my answer, if my answer is wrong please clarify.?

Question

my answer is the PH of 1.40
here is the question
calculate the ph of a solution made by combining 200.00ml of 0.235M chlorous acid(Ka=1.2E-2) with 200.00ml of 0.200M hydrocyanic acid(Ka=6.2E-10) and 300.00ml of 0.150M ammonium chloride(Ka=5.6E-10)
if i am right just say i am right if i am wrong give me the heads up.

the method i use
add all Ka
find out the final M
use the pKa+log(base/acid)

are you sure?

so is ph= 3.1 then

Answer 1

No, it is not correct.
Also you can't do the calculations separately and add [H+].
You have 3 weak acids, so the dissociation of one will effect that of the others.

First find your final volume Vf=200+200+300 =700 ml
Then find the new concentration using M1*V1=M2*V2
For HClO2 0.235*200= M2*700 => M2=0.067
For HCN 0.200*200=M2*700 => M2=0.057
For NH4Cl 0.150*300=M2*700 => M2=0.064

You see that the 3 acids have approximately the same concentration, while HClO2 has a Ka 8 orders of magnitude higher than the other 2. So it would be safe to assume that the contribution of the other acids is 0 and calculate the pH as if you only have 0.067 M HClO2

.. .. .. .. .. HClO2 <=> H(+) + ClO2(-)
initial .. .. ..0.067
dissoc. .. .. x
Produce .. .. .. .. .. .. .. x .. .. .. .. x
At equil .. 0.067-x .. .. ..x .. .. .. .. x

Ka=x^2/(0.067-x)

approx 0.067 >> x and 0.067-x=0.067
thus Ka=x^2/0.067=>

x= sqrt(0.067*0.012)=0.028.

It is NOTmuch smaller than 0.067 so you HAVE TO solve the quadratic

0.012=x^2/(0.067-x)=>
x^2+0.012x-8.04*10^-4=0 =>

x=0.023

pH=-logx =-log(0.023) =1.64

If you want to find the exact value, then you have to set up 3 ICE tables. If x mole/l HClO2 dissociate, y for HCN and z for NH4Cl,
then you have at equilibrium [H+]=x+y+z

Ka(HClO2)= x(x+y+z)/ (0.067-x)=0.012
Ka(HCN)= y(x+y+z)/(0.057-y)=6.2*10^-10
Ka(NH4Cl)= z(x+y+z)/(0.064-z)=5.6*10^-10

You have a system of equations that if you solve the only acceptable set of solutions is
x=0.023 (as we calculated before)
y= 1.54*10^-9
z=1.56*10^-9

So you see that y and z are so small that they can be neglected.

Answer 2

I don't think you can add Ka values, and the solution is not a buffer.

Work out the H+ concentration for each of the three substances by using [H+] = root (Ka x molarity)

Then add all the [H+] concentrations, and then take the negative logarithm to get the pH.