calculate the change in pH when 9.00ml of 0.100M HCl(aq) is added to 100.0mL of a buffer solution that is 0.100M in NH3(aq) and 0.100M NH4Cl(aq).
...Calculate the change in pH when 9.00,L if 0.100M NaOH(aq) is added to the original buffer solution.
Here is how you do these; I'll do the first one.
9.00 mL x 0.100M HCl = 0.900 mmols.
100.0 mL x 0.100 M NH3 = 10.0 mmols.
100.0 mL x 0.100 M NH4Cl = 10.0 mmols.
...........NH3 + H^+ ==> NH4^+
I.........10.....0.......10
add.............0.9..............
C.......-0.9...-0.9......+0.9
E.........9.1...0.........10.9
pH = pKa + log(base)/(acid)
What are you using for pKa? Use the one in your text and not what I've used.
pH = 9.24 + log(9.1/10.9) = 9.16.
The question asks for CHANGE in pH. It was 9.24 before the HCl was added. Change is 9.24 to 9.16 = -0.08. Check my work. Watch the significant figures; I didn't.
Why did the chemist get a dog?
Because he wanted a pet with a good pH balance!
Now, let's get back to your question. To determine the change in pH when 9.00 mL of 0.100 M HCl(aq) is added, we need to consider the reaction between HCl and NH3. HCl is a strong acid, so it will completely dissociate, while NH3 is a weak base.
First, let's calculate the moles of HCl added:
moles HCl = 0.100 M x 0.00900 L = 0.0009 mol
Since 1 mole of HCl reacts with 1 mole of NH3, the amount of NH3 that reacts will also be 0.0009 mol.
Next, let's calculate the new concentrations of NH3 and NH4Cl in the buffer solution after the reaction:
NH3: 0.0009 mol / (0.100 L + 0.009 L) = 0.0078 M
NH4Cl: 0.100 M + 0.0009 mol / (0.100 L + 0.009 L) = 0.109 M
Now, let's calculate the pOH using the new NH3 concentration:
pOH = -log(0.0078) ≈ 2.11
Finally, we can calculate the change in pH by using the formula pH + pOH = 14:
change in pH = 14 - pOH ≈ 14 - 2.11 ≈ 11.89
So, the change in pH when 9.00 mL of 0.100 M HCl(aq) is added to the buffer solution is approximately 11.89.
Now, let's answer the second part of your question about the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
If you add NaOH to the solution, it will react with the NH4Cl to form NH3 and water. This will result in a decrease in the concentration of NH4Cl and an increase in the concentration of NH3.
To calculate the moles of NaOH added:
moles NaOH = 0.100 M x 0.00900 L = 0.0009 mol
Since 1 mole of NaOH reacts with 1 mole of NH4Cl, the amount of NH4Cl that reacts will also be 0.0009 mol.
Next, let's calculate the new concentrations of NH3 and NH4Cl in the buffer solution after the reaction:
NH3: 0.100 M + 0.0009 mol / (0.100 L + 0.009 L) = 0.109 M
NH4Cl: 0.100 M - 0.0009 mol / (0.100 L + 0.009 L) = 0.091 M
Using the new NH3 concentration, let's calculate the new pOH:
pOH = -log(0.109) ≈ 0.961
Finally, we can calculate the change in pH by using the formula pH + pOH = 14:
change in pH = 14 - pOH ≈ 14 - 0.961 ≈ 13.04
So, the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution is approximately 13.04.
Phew, that was quite a chemical balancing act!
To calculate the change in pH when 9.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M NH4Cl(aq), we need to use the Henderson-Hasselbalch equation and consider the dissociation of NH3 and NH4Cl.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pH is the solution's pH, pKa is the acid dissociation constant (the negative logarithm of the Ka), [A-] is the concentration of the base, and [HA] is the concentration of the acid.
1. Calculate the initial pH of the buffer solution:
To apply the Henderson-Hasselbalch equation, we need the pKa value for the NH4Cl/NH3 buffer system. The pKa for NH4+ is about 9.25. Therefore, pKa = 9.25.
pH = pKa + log([A-]/[HA])
pH = 9.25 + log([NH3]/[NH4+])
pH = 9.25 + log(0.100/0.100)
pH = 9.25 + log(1)
pH = 9.25 + 0
pH = 9.25 (initial pH of the buffer solution)
2. Calculate the moles of HCl added to the buffer solution:
Using the equation C1V1 = C2V2, where C1 and V1 are the concentration and volume of the initial solution, and C2 and V2 are the concentration and volume after adding the HCl.
C1V1 = C2V2
(0.100 M)(9.00 mL) = (C2)(100.0 mL)
C2 = (0.100 M)(9.00 mL) / (100.0 mL)
C2 = 0.009 M (concentration of HCl in the mixture)
moles of HCl = (concentration)(volume) = (0.009 M)(0.009 L) = 0.000081 mol
3. Determine the final concentrations of NH3 and NH4+ after adding HCl:
The reaction between HCl and NH3 results in the formation of NH4Cl.
Since the ratio of NH3 to NH4Cl is 1:1, the final concentrations of NH3 and NH4+ will both be equal to the initial concentration of NH3 minus the concentration of HCl added:
[NH3] = [NH3]initial - [HCl]
[NH3] = 0.100 M - 0.009 M
[NH3] = 0.091 M
[NH4+] = [NH3] = 0.091 M
4. Calculate the final pH after adding HCl:
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 9.25 + log([NH3]/[NH4+])
pH = 9.25 + log(0.091/0.091)
pH = 9.25 + log(1)
pH = 9.25 + 0
pH = 9.25 (final pH after adding HCl)
The change in pH after adding 9.00 mL of 0.100 M HCl(aq) to the original buffer solution is 0.00 (no change in pH).
To calculate the change in pH when 9.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution, we will repeat the calculations:
1. Calculate the initial pH of the buffer solution:
pH = 9.25 (from previous calculations)
2. Calculate the moles of NaOH added to the buffer solution:
Using the equation C1V1 = C2V2, where C1 and V1 are the concentration and volume of the initial solution, and C2 and V2 are the concentration and volume after adding NaOH.
C1V1 = C2V2
(0.100 M)(9.00 mL) = (C2)(100.0 mL)
C2 = (0.100 M)(9.00 mL) / (100.0 mL)
C2 = 0.009 M (concentration of NaOH in the mixture)
moles of NaOH = (concentration)(volume) = (0.009 M)(0.009 L) = 0.000081 mol
3. Determine the final concentrations of NH3 and NH4+ after adding NaOH:
Since NaOH is a strong base, it will react with NH4+ to form NH3 and water (H2O). The moles of NH4+ is equal to the moles of NaOH added:
[NH4+] = [NH4+]initial - [NaOH]
[NH4+] = 0.100 M - 0.009 M
[NH4+] = 0.091 M
[NH3] remains unchanged at 0.091 M.
4. Calculate the final pH after adding NaOH:
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 9.25 + log([NH3]/[NH4+])
pH = 9.25 + log(0.091/0.091)
pH = 9.25 + log(1)
pH = 9.25 + 0
pH = 9.25 (final pH after adding NaOH)
The change in pH after adding 9.00 mL of 0.100 M NaOH(aq) to the original buffer solution is 0.00 (no change in pH).
To calculate the change in pH when a solution is added to a buffer, we need to consider the effect of the added species on the equilibria of the buffer components.
For the first part of the question, we are adding HCl to the buffer solution. When HCl is added to water, it fully dissociates to H+ and Cl-. The H+ ions will react with the NH3 in the buffer, causing a change in the equilibrium of the buffer system. The overall reaction can be represented as:
NH3(aq) + H+(aq) ↔ NH4+(aq)
In this reaction, NH3 is acting as a base and the H+ ions from HCl are acting as an acid. As a result, the NH4+ concentration will increase and the NH3 concentration will decrease.
To calculate the change in pH, we need to calculate the concentrations of NH3 and NH4+ after the reaction occurs. The initial concentration of NH3 is given as 0.100M, and the initial concentration of NH4+ is also 0.100M (since it is from the buffer solution).
First, let's calculate the moles of H+ ions added to the solution.
Moles of HCl = concentration of HCl * volume of HCl added
Moles of HCl = 0.100M * 9.00mL = 0.900mmol
Since 1 mole of HCl dissociates into 1 mole of H+, in this case, we have 0.900mmol of H+ added to the buffer solution.
Next, we need to consider the reaction between H+ and NH3. From the balanced equation, we know that the ratio between H+ and NH3 is 1:1. Therefore, the moles of NH3 consumed will be equal to the moles of H+ added. In this case, 0.900mmol of NH3 will be consumed.
Now, let's calculate the new concentrations of NH3 and NH4+ after the reaction. Since the total volume of the solution is 100.0mL + 9.00mL = 109.00mL, we need to convert the moles to concentrations.
Concentration of NH3 = (moles of NH3 remaining) / (total volume of solution)
Concentration of NH3 = (0.100M * 0.100L - 0.900mmol) / 0.109L
Concentration of NH3 = 0.0826M
Concentration of NH4+ = (moles of NH4+ formed) / (total volume of solution)
Concentration of NH4+ = (0.100M * 0.100L + 0.900mmol) / 0.109L
Concentration of NH4+ = 0.183M
Now, we can calculate the pOH of the solution using the new concentration of NH3:
pOH = -log10(NH3 concentration)
pOH = -log10(0.0826) = 1.08
Finally, we can calculate the pH of the solution:
pH = 14 - pOH
pH = 14 - 1.08 = 12.92
The change in pH is the difference between the initial pH of the buffer solution and the pH after the reaction. In this case, the change in pH is 12.92 - 14 = -1.08.
For the second part of the question, when NaOH is added to the buffer solution, the OH- ions will react with the NH4+ in the buffer, causing a change in the equilibrium of the buffer system. The overall reaction can be represented as:
NH4+(aq) + OH-(aq) ↔ NH3(aq) + H2O(l)
In this reaction, NH4+ acts as an acid and the OH- ions from NaOH act as a base. As a result, the NH3 concentration will increase, and the NH4+ concentration will decrease.
The approach to calculating the change in pH is similar to the first part. First, calculate the moles of OH- ions added by NaOH. Then, determine the moles consumed for both NH4+ and OH- based on the stoichiometric ratio. Finally, calculate the new concentrations of NH4+ and NH3 and use them to determine the new pOH and pH of the solution.
I hope this helps! If you have any further questions, please, let me know.