How to Calculate Species Concentrations in a 0.010 M Solution of NaN3?

[lha08]

Homework Statement

Soldium azide, NaN3 is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for hydrazoic acid, HN3 is 1.9X10^-5.

Homework Equations

The Attempt at a Solution

I'm sort of confused because in the beginning i know that i have to separate NaN3 into its component elements..so in this case..NaN3-->Na+N3? or is it NaN3-->Na+3N?

 

[Borek]

NaN₃ is a salt of a weak acid HN₃.

 

[Blueshift5]

Firstly, it is important to clarify that NaN3 (sodium azide) is a salt, not an acid or a base. However, when dissolved in water, it can produce a weak acid, hydrazoic acid (HN3). Therefore, the concentration of all species in a 0.010 M solution of NaN3 can be calculated by considering the dissociation of HN3.

The balanced equation for the dissociation of HN3 in water is:

HN3 + H2O ↔ H3O+ + N3-

From this equation, we can see that for every 1 mole of HN3 that dissociates, we will get 1 mole of H3O+ and 1 mole of N3-. Therefore, the concentration of H3O+ and N3- will be equal to the concentration of HN3.

To calculate the concentration of HN3, we can use the Ka value given in the problem. The Ka expression for HN3 is:

Ka = [H3O+][N3-]/[HN3]

Using the given Ka value of 1.9x10^-5 and the concentration of HN3 (which is equal to the concentration of NaN3), we can solve for the concentration of H3O+ and N3-:

1.9x10^-5 = [H3O+][N3-]/0.010

[H3O+][N3-] = 1.9x10^-7

Since [H3O+] = [N3-] (as discussed above), we can substitute [H3O+] = [N3-] = x into the equation:

x^2 = 1.9x10^-7

Solving for x, we get x = [H3O+] = [N3-] = 1.38x10^-4 M.

Therefore, the concentration of all species in a 0.010 M solution of NaN3 is:

[H3O+] = [N3-] = 1.38x10^-4 M

[NaN3] = 0.010 M

[Na+] = 0.010 M

I hope this helps clarify any confusion. It is always important to carefully read and understand the problem before attempting to solve it.