Calculate the end point pH, when 25 mL of 0.01 mol/L HCl solution reacts exactly with 25 mL of 0.1 mol/L NH4OH solution.
NH3 Kb = 1.8 x 10^-5
This will be the pH of NH4Cl solution.
NH4+ + HOH ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+). Solve for H3O^+, then pH = - log(H3O^+). Post your work if you get stuck.
But we haven't learned how to get the pH of H3O+...we only get the pH of [H+].
Same thing. They are all H3O^+ but sometimes we get a little lazy and instead of writing H3O^+ or H*H2O^+ we just omit the water part and write H^+.
Ah, pH calculations can be a bit tricky, but don't worry, I'm here to help you out! Let's jump right in and calculate the end point pH.
First, let's find out how many moles of HCl and NH4OH we have. We can use the formula n = C x V, where n is the number of moles, C is the concentration, and V is the volume.
For HCl:
n(HCl) = 0.01 mol/L × 0.025 L = 0.00025 mol
And for NH4OH:
n(NH4OH) = 0.1 mol/L × 0.025 L = 0.0025 mol
Now, let's determine which reagent is the limiting one. Since the reaction is 1:1 between HCl and NH4OH, we can see that HCl is the limiting reagent because it has fewer moles.
The balanced chemical equation for the reaction is:
HCl + NH4OH → NH4Cl + H2O
So, all the HCl will react with NH4OH to form NH4Cl and water. Since NH4Cl is a salt, it will dissociate in water. In this case, NH4Cl will dissociate to form NH4+ and Cl- ions.
Now, let's look at the equation for the dissociation of NH4+ in water:
NH4+ + HOH → NH3 + H3O+
According to the equation, NH4+ acts as an acid and donates a proton to water to form NH3 and H3O+.
To find the concentration of H3O+ ions, we can use the equation:
Ka = Kw/Kb = (NH3)(H3O+)/(NH4+)
Given that Kb for NH3 is 1.8 × 10^-5, we can calculate Ka:
Ka = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.56 × 10^-10
Now, let's rearrange the equation and solve for the concentration of H3O+:
(H3O+) = (NH4+) × (Ka)/(NH3) = (0.00025 mol)/(0.0025 mol) × (5.56 × 10^-10)
(H3O+) = 0.0001 × 5.56 × 10^-10 = 5.56 × 10^-14 mol/L
Finally, let's calculate the pH using the formula pH = -log(H3O+):
pH = -log(5.56 × 10^-14)
pH ≈ 13.3
So, the end point pH of the NH4Cl solution is approximately 13.3. Keep in mind that this result is based on the assumptions we made and the given values.
To calculate the end point pH, we need to first determine the concentration of H3O+ ions in the solution, which will then allow us to calculate the pH.
Let's start by writing the balanced chemical equation for the reaction between HCl and NH4OH:
HCl + NH4OH -> NH4Cl + H2O
Since the reaction is between a strong acid (HCl) and a weak base (NH4OH), we can assume that the HCl completely dissociates and the NH4OH partially dissociates.
The balanced equation for the partial dissociation of NH4OH is:
NH4OH + H2O -> NH4+ + OH-
Given that the initial concentration of NH4OH is 0.1 mol/L and the volume is 25 mL, we can calculate the number of moles of NH4OH:
moles of NH4OH = concentration x volume
= 0.1 mol/L x 0.025 L
= 0.0025 mol
Since the reaction is stoichiometrically balanced, the moles of NH4+ formed will also be 0.0025 mol.
Now, we need to consider the reaction between NH4+ and water, which forms NH3 and H3O+:
NH4+ + H2O -> NH3 + H3O+
Since NH4+ and NH3 are a conjugate acid-base pair, we can use the ionization constant (Ka) to determine the concentration of H3O+.
Given that Ka = Kw / Kb, and Kb for NH3 is 1.8 x 10^-5, we can rearrange the equation to solve for Ka:
Ka = Kw / Kb
Ka = (1.0 x 10^-14) / (1.8 x 10^-5)
Ka = 5.56 x 10^-10
Using the balanced equation, we can write an expression for Ka:
Ka = ([NH3][H3O+]) / [NH4+]
At the end point, when the reaction is complete, the concentration of NH4+ will be equal to the moles of NH4+ divided by the volume of the solution:
[NH4+] = moles of NH4+ / volume
= 0.0025 mol / 0.05 L
= 0.05 mol/L
Now we can substitute the values into the expression for Ka and solve for [H3O+]:
5.56 x 10^-10 = ([NH3][H3O+]) / 0.05 mol/L
From here, we can isolate [H3O+]:
[H3O+] = (5.56 x 10^-10) * 0.05 mol/L / [NH3]
Now, we need to determine the concentration of NH3, which is equal to the initial concentration of NH4OH minus the amount that has reacted. At the end point, all the NH4OH has reacted, so there is no NH3 left. Therefore, [NH3] = 0 mol/L.
Substituting this value into the equation for [H3O+], we get:
[H3O+] = (5.56 x 10^-10) * 0.05 mol/L / 0 mol/L
Since we cannot divide by zero, we can conclude that at the end point, there is no NH3 left in the solution, and therefore, the concentration of H3O+ is 0.
Since the concentration of H3O+ is 0, the pH of the solution at the end point is infinite or undefined.
Therefore, the end point pH is not calculable in this particular scenario.
Note: In a real experimental setup, it might not be possible to reach complete reaction and remove all traces of NH4+ or NH3 from the solution, which is why the pH value might differ from being infinite or undefined.
To calculate the end point pH, we need to first determine the concentration of H3O+ ions in the NH4Cl solution. We can do this by using the equation for the ionization of NH4+ and H2O, and the equilibrium constant expression, Ka.
The equation for the ionization reaction is:
NH4+ + H2O ⇌ NH3 + H3O+
From the equation, we can see that the concentration of NH3 and H3O+ ions will be equal to each other since the reaction is stoichiometrically balanced.
The equilibrium constant expression, Ka, is defined as:
Ka = [NH3][H3O+] / [NH4+]
Since the concentration of NH3 and H3O+ is the same, we can substitute [NH3] with [H3O+] in the equation:
Ka = [H3O+]^2 / [NH4+]
Given that the Kb value for NH3 is 1.8 x 10^-5, we can use the relationship between Ka and Kb:
Ka = Kw / Kb
where Kw is the equilibrium constant for water, which is 1.0 x 10^-14 at 25°C.
Substituting the values, we can solve for Ka:
1.0 x 10^-14 = Ka * Kb
Ka = 1.0 x 10^-14 / Kb
= 1.0 x 10^-14 / 1.8 x 10^-5
≈ 5.56 x 10^-10
Now that we have the value of Ka, we can solve for [H3O+]:
Ka = [H3O+]^2 / [NH4+]
[H3O+]^2 = Ka * [NH4+]
[H3O+]^2 = (5.56 x 10^-10) * (0.01 mol/L)
[H3O+]^2 ≈ 5.56 x 10^-12
[H3O+] ≈ √(5.56 x 10^-12)
[H3O+] ≈ 7.46 x 10^-6 mol/L
Finally, the end point pH can be calculated by taking the negative logarithm (base 10) of [H3O+]:
pH = -log10[H3O+]
= -log10(7.46 x 10^-6)
≈ 5.13
Therefore, the end point pH of the NH4Cl solution is approximately 5.13.