A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before
the addition of any KOH. The Kb of NH3 is 1.8 × 10-5.
I thought that I would need a second volume to find the moles of NH3?
Something is wrong with the post. You're titrating NH3 with HNO3 but you want the pH before any KOH is added. Where did the KOH come from? I assume you want the pH of a solution of 0.1M NH3.
At the beginning, before any HNO3 has been added
...........NH3 + HOH ==> NH4^+ + OH^-
I..........0.1............0.......0
C...........-x............x.......x
E..........0.1-x..........x.......x
Substitute the E line into the Kb expression and solve for x = (OH^-), then convert that to pH.
Well, isn't it refreshing to find someone who appreciates the importance of having a second volume in life! But in this case, you're in luck because you don't actually need a second volume to find the moles of NH3. Since you know the initial concentration of NH3 (0.10 M) and the volume of the solution (100.0 mL), you can use the good ol' formula C = n/V to find the number of moles of NH3. Don't worry, this formula won't bite... unless it's a venomous equation! *wink*
Just plug in the values and calculate the moles of NH3. Once you know that, you can move forward with calculating the pH of the solution before the addition of any KOH. And hey, if you need any more help along the way, I'm here to add a dash of humor to make chemistry a little less... element-ary!
Yes, you are correct. In order to determine the pH of the solution before the addition of any KOH, you will need to calculate the concentration of NH3 (ammonia) in the solution.
To find the concentration of NH3, you can use the equation:
NH3 + H2O ⇌ NH4+ + OH-
The concentration of NH3 can be calculated using the Kb value, which is the equilibrium constant for the reaction:
Kb = [NH4+][OH-]/[NH3]
Since you know the Kb value is 1.8 × 10^-5, you can convert this into Kw (the water constant) by using Kw = Ka * Kb, where Ka is the acid dissociation constant of water (1.0 × 10^-14).
Kw = Ka * Kb
1.0 × 10^-14 = (x)(1.8 × 10^-5)
By solving this equation, you can find the concentration of OH- ions in the solution. However, the concentration of OH- is equal to the concentration of NH4+ (NH4OH is a weak base), so you will have the concentration of NH4+ as well.
Now that you have the concentration of NH4+ and OH-, you can find the concentration of NH3 by using the equation:
NH3 + H2O ⇌ NH4+ + OH-
0.10 M - x + x = [NH3]
Finally, you can calculate the pOH of the solution using the concentration of OH- ions. The pOH is the negative logarithm of the OH- concentration:
pOH = -log[OH-]
Once you have the pOH, you can find the pH using the equation:
pH = 14 - pOH
Remember to use the initial concentration of NH3 (0.10 M) before the addition of any KOH.
To determine the pH of the solution before the addition of any KOH, you can use the concept of the common-ion effect. This effect occurs when a solution containing a weak base and its conjugate acid is titrated with a strong acid.
In this case, NH3 (ammonia) is a weak base, and HNO3 is a strong acid. Before the addition of any KOH, only the NH3 and HNO3 are present. The NH3 will react with the HNO3 to form NH4+ (ammonium) and NO3- (nitrate) ions.
To calculate the pH, we need to analyze the equilibrium between NH3 and NH4+ ions. The equilibrium equation is:
NH3 + H2O ⇌ NH4+ + OH-
The Kb of NH3 is given as 1.8 × 10^-5, which is the equilibrium constant for the reaction:
NH3 + H2O ⇌ NH4+ + OH-
The Kb expression for this reaction can be written as:
Kb = [NH4+][OH-] / [NH3]
Since the NH3 concentration is given as 0.10 M and the NH4+ concentration is initially zero, we can assume that the initial concentration of OH- is also zero. Therefore, the equation simplifies to:
Kb = [OH-] / [NH3]
Now, we can solve for the concentration of OH- using the Kb value and the initial concentration of NH3:
1.8 × 10^-5 = [OH-] / 0.10
Solving for [OH-]:
[OH-] = 1.8 × 10^-5 × 0.10
[OH-] = 1.8 × 10^-6
Now, using the fact that [H+] × [OH-] = 1.0 × 10^-14 (at 25°C), we can solve for [H+], which is the concentration of hydrogen ions:
[H+] = 1.0 × 10^-14 / [OH-]
Substituting the value of [OH-]:
[H+] = 1.0 × 10^-14 / 1.8 × 10^-6
[H+] ≈ 5.6 × 10^-9
Finally, we can calculate the pH using the formula:
pH = -log[H+]
pH ≈ -log(5.6 × 10^-9)
pH ≈ 8.25
Therefore, the pH of the solution before the addition of any KOH is approximately 8.25.