What is the pH of the solution ?

Question

5.23 g of hydrazoic acid ( HN3 ) is added to water to make a final solution of 408 ml.
What is the pH of the solution ?

Use Ka = 1.90 x 10-5 at 25.0oC.

Answer 1

Find the moles of HN3 (BTW, this is not a stable compound) by dividing the 5.23 g by it mol wt.

Divide that by 0.408 to find molarity. Call that M.

Now, use the equilibrium expression :
[azide ion][H+]/[HN3] = 1.9x10-5
Let x be the amount of HN3 dissociated. As a first appx, assume x< x^2/M= 1.9x10-5, and x = sqrt(M*1.9x10-5)appx.
x is also the [H+], so you can figure pH from x.