Calculate the pH of 0.15M hydrazoic acid, HN3
. Ka = 1.8 x 10-5
.
...............HN3 ==> H^+ + N^3-
I..............0.15..........0.........0
C...............-x............x..........x
E..............0.15-x......x...........x
Plug the E line (equilibrium line) into the Ka expression and solve for x = (H^+).
Then convert H^+ to pH by pH = - log (H^+). Post your work if you get stuck.
To calculate the pH of hydrazoic acid (HN3), we need to consider the dissociation of the acid into its ions. In this case, hydrazoic acid would dissociate into a hydrazoate ion (N3-) and a hydrogen ion (H+).
The first step is to write down the balanced equation for the dissociation reaction of hydrazoic acid:
HN3 ⇌ H+ + N3-
From the balanced equation, we can see that for every 1 mole of hydrazoic acid (HN3) that dissociates, 1 mole of hydrogen ions (H+) is produced. Therefore, the concentration of hydrogen ions (H+) will be equal to the concentration of the hydrazoic acid (HN3).
Given that the concentration of hydrazoic acid (HN3) is 0.15 M, the concentration of hydrogen ions (H+) is also 0.15 M.
The pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+). Therefore, to find the pH, we need to calculate the logarithm of the concentration of hydrogen ions and then take the negative value.
Using the concentration of hydrogen ions (H+) = 0.15 M, we can calculate the pH by using the formula:
pH = -log[H+]
Substituting the value of the concentration of hydrogen ions into the formula:
pH = -log(0.15)
Using a scientific calculator:
pH ≈ 0.82
Therefore, the pH of 0.15 M hydrazoic acid (HN3) is approximately 0.82.