4.54 g of hydrazoic acid ( HN3 ) is added to water to make a final solution of 504 ml.
What is the pH of the solution ?
Use Ka = 1.90 × 10−5
HN3 + H2O ==> H3O^+ + N3^-
Convert 4.54 grams to moles and divide by volume in liters to obtain molarity. Set up an IC# chart and calculate H^+, then convert to pH.
To find the pH of the solution, we need to determine the concentration of hydrazoic acid (HN3) in the solution.
We are given that 4.54 g of HN3 is added to water to make a final solution of 504 ml. We can start by converting the given mass of HN3 to moles.
The molar mass of HN3 is:
(1 × 1.00784 g/mol) + (1 × 14.00674 g/mol) + (3 × 15.999 g/mol) = 43.02674 g/mol
So, 4.54 g of HN3 is equal to:
4.54 g ÷ 43.02674 g/mol = 0.10577071 mol
Next, we need to calculate the concentration of HN3 in the solution.
Given that the final solution has a volume of 504 ml, we can convert this to liters:
504 ml ÷ 1000 ml/L = 0.504 L
Now, we can calculate the concentration (C) in mol/L:
C = moles/volume = 0.10577071 mol / 0.504 L = 0.2098 mol/L
Since hydrazoic acid is a weak acid, we can calculate the concentration of H+ ions using the acid dissociation constant (Ka).
In the equation HN3 ⇌ H+ + N3-, we can assume that the reaction goes to completion, so the concentration of H+ ions would be equal to the concentration of HN3.
Now, we can calculate the pH of the solution using the concentration of H+ ions:
pH = -log[H+]
pH = -log(0.2098)
Using a calculator or logarithmic tables, we can calculate the pH to be 0.679.