Good at chem problems?? I will give you 10pts if you help me find the answer!!!?

Question

(HN3) (3=subscript)is a low boiling point liquid that is soluble in water in small quantities. It partially dissociates to form hydronium ions and azide ion (N3-)(3=subscript), with the pKa of 1.9 x 10-5 (5=subscript).

Calculate the pH of a solution of 0.724 g of HN3 in 400mL of water. Ignore any increase in volume due to small amount of HN3.

show work please.

Answer 1

Let's assume the concentration of your acid is C.
Then

.. .. .. .. .. .. HN3 <=> H(+) + N3(-)
Initial .. .. .. .. C
Dissociate. .. x
Produce .. .. .. .. .. .. .. x .. .. .. x
At Equil. .. .. C-x .. .. .. x .. .. .. x

Ka= [H+][N3-]/[HN3]= x^2/(C-x)

So let's find out the value of C
C=mole/V
and since mole=mass/MW
C=mass/(MW*V) =0.724/(43*0.4) =0.042 M
So the equation becomes Ka= x^2/(0.042-x)= 1.9*10^-5

Let's assume that 0.042 >> x. Then practicalyl 0.042-x=0.042 and the equation becomes x^2/0.042= 1.9*10^-5 =>
x= Squareroot (0.042*1.9*10^-5) = 0.000893 << 0.042 so our assumption is fair (otherwise you would solve the quadratic equation for an exact solution)

pH=-logx=-log(8.93*10^-4)=3.05

Answer 2

I think you probably mean NH3...
The pKa means that 10 to that power is your equilibrium value.
i.e. 10^(1.9x10^-5)=[H-]^3*[N3-]/[NH3]
A little mathematics will do you good!

Answer 3

Ka = [NH3][H3O+]/[NH4+]= 5.60 x 10-10
[NH3] = [H3O+]
0,724g NH3*(1mol NH3/17g NH3)= 0,0426 moles NH3
0,0426 moles / 0,400L = 0,106 M
0,106M NH4+
5.60 x 10-10 = [H3O+]^2/0,106
[H3O+] = 7,70x10^-6
pH = -log[H+] = -log[H3O+]
-log[H3O+]= 5,11
pH = 5,11