Determine the PH of a solution that results from mixing 50 ml of .200 M hydroazoic acid, HN3 with 50 ml of CA(OH)2
You don't need M Ca(OH)2??? I don't think you can work the problem without it. .
sorry that's .2 M caoh2
You meant Ca(OH)2.
This is new to me but I believe this is an acid/base reaction between HN3 and Ca(OH)2. I think the equation is
2HN3 + Ca(OH)2 ==> Ca(N3)2 + 2H2O
mols HN3 = M x L = 0.2 x 0.05 = 0.01
mols Ca(OH)2 = M x L = 0.2 x 0.05 = 0.01
We start with this probably being a limiting reagent problem. I work those the long way. First, how much Ca(N3)2 could be formed from 0.01 mol Ca(OH)2? That will be 0.01 x [1 mol Ca(N3)2/1 mol Ca(OH)2] = 0.01 mol Ca(N3)2
How much Ca(N3)2 could be formed from 0.01 mol HN3? That's 0.01 mol HN3 x (1 mol Ca(N3)2/2 mol HN3) = 0.01 x 1/2 = 0.005 mols Ca(N3)2; therefore, HN3 is the limiting reagent and you will have 0.005 mols of the salt formed, all of the HN3 will be used and you will have 0.005 mols Ca(OH)2 in excess. Convert that to M = 0.005 mols/0.1L = 0.05M
Now all you need to do is to convert 0.05M Ca(OH)2 to pH. The HN3 has a Ka of 1.9E-5; therefore, it will contribute negligible OH or H.
To determine the pH of the resulting solution, you need to consider the reaction that occurs between hydroazoic acid (HN3) and calcium hydroxide (Ca(OH)2). The reaction can be written as follows:
HN3 + Ca(OH)2 → N3- + Ca+2 + 2OH-
In this reaction, hydroazoic acid reacts with calcium hydroxide to form azide ion (N3-) and calcium ion (Ca+2), along with two hydroxide ions (OH-).
To determine the pH, we need to consider how these ions influence the acidity/basicity of the solution. The hydroxide ions (OH-) are a strong base and will increase the pH, while the azide ion (N3-) is a weak base and will also increase the pH but to a lesser extent. The calcium ion (Ca+2) has no effect on the pH because it is a spectator ion and does not participate in the acidity/basicity of the solution.
To determine the concentration of hydroxide ions (OH-) in the final solution, we need to calculate how many moles of each reactant are present.
Calculating the moles of hydroazoic acid (HN3):
Moles of HN3 = volume (in L) × molarity
= 0.050 L × 0.200 mol/L
= 0.010 mol
Calculating the moles of calcium hydroxide (Ca(OH)2):
Moles of Ca(OH)2 = volume (in L) × molarity
= 0.050 L × 0.200 mol/L
= 0.010 mol
From the balanced equation, we see that 1 mole of hydroazoic acid produces 2 moles of hydroxide ions (OH-). So, the moles of hydroxide ions formed will be twice the moles of hydroazoic acid used.
Moles of OH- = 2 × 0.010 mol
= 0.020 mol
Now, to calculate the concentration of hydroxide ions in the final solution, we divide the moles of OH- by the total volume of the resulting solution (100 mL or 0.100 L). Keep in mind that 0.050 L + 0.050 L = 0.100 L.
Concentration of OH- = moles of OH- / total volume
= 0.020 mol / 0.100 L
= 0.200 M
Therefore, the concentration of hydroxide ions (OH-) in the final solution is 0.200 M.
Since the hydroxide ion concentration is known, we can now calculate the pOH of the solution using the formula:
pOH = -log[OH-]
pOH = -log(0.200)
pOH ≈ 0.70
Finally, to find the pH, we use the relation:
pH = 14 - pOH
pH = 14 - 0.70
pH ≈ 13.30
Therefore, the pH of the solution resulting from mixing 50 mL of 0.200 M hydroazoic acid with 50 mL of Ca(OH)2 is approximately 13.30.