To 160.0 mL of a buffer solution containing 0.250 M ammonia and 0.100 M ammonium chloride, 20.00 mL of 1.00M HClO4 are added. a) Calculate the pH of the new solution. b) What becomes the pH if another 20.00 mL of the same acid solution is added?
a) Well, well, well, looks like we have a bit of a chemistry puzzle to solve here! Let's get cracking, shall we?
First things first, we need to find out what happens when we add that HClO4 to our buffer solution. The ammonia (NH3) will react with the HClO4 and form NH4+ and ClO4-. The NH4+ will then react with the NH3, forming NH4Cl and restoring the buffer equilibrium.
Now, to calculate the new pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, the ammonia (NH3) is the base (A-) and the ammonium chloride (NH4Cl) is the acid (HA). The pKa value for the ammonium ion (NH4+) is 9.25.
Let's plug in the numbers: the concentration of NH3 is 0.250 M, and the concentration of NH4Cl is 0.100 M. So,
pH = 9.25 + log (0.100/0.250)
= 9.25 + log (0.4)
= 9.25 + (-0.3979)
= 8.8521
So, the pH of the new solution is approximately 8.85.
b) Now, if we add another 20.00 mL of the same acid solution (HClO4), we're simply doubling the amount of H+ ions in the solution. And by golly, that's going to have an impact on our pH!
Let's use the same Henderson-Hasselbalch equation to calculate the new pH. In this case, we'll need to recalculate the concentration of the acid (HA):
HA = (volume x concentration) + (volume x concentration)
= (160 mL x 0.100 M) + (40 mL x 1.00 M)
= 16 mL + 40 mL
= 56 mL
The total volume of the solution is now 180 mL.
pH = 9.25 + log ([A-]/[HA])
= 9.25 + log (0.100/0.280)
≈ 9.25 + (-0.5777)
≈ 8.67
So, if we add another 20.00 mL of the same acid solution, the pH of the solution would be approximately 8.67.
Hope this chemical circus act was entertaining!
To solve this problem, we will first calculate the moles of ammonia and ammonium chloride present in the initial buffer solution. Then, we will determine the effect of adding 20.00 mL of 1.00 M HClO4 on the pH of the solution. Finally, we will repeat the calculations after adding another 20.00 mL of the same acid solution to find the new pH.
Let's start with part (a) and calculate the pH after adding 20.00 mL of 1.00 M HClO4.
Step 1: Calculate moles of ammonia and ammonium chloride
Moles of ammonia (NH3) = Volume (L) × Concentration (mol/L)
Moles of ammonia = 0.160 L × 0.250 mol/L = 0.04 mol
Moles of ammonium chloride (NH4Cl) = Volume (L) × Concentration (mol/L)
Moles of ammonium chloride = 0.160 L × 0.100 mol/L = 0.016 mol
Step 2: Determine the limiting reactant
The limiting reactant is the one with the smallest number of moles. In this case, ammonium chloride (NH4Cl) has fewer moles, so it is the limiting reactant.
Step 3: Calculate the moles of excess HClO4
Moles of HClO4 = Volume (L) × Concentration (mol/L)
Moles of HClO4 = 0.020 L × 1.00 mol/L = 0.020 mol
Step 4: Determine the moles of NH4+ ions formed
Moles of NH4+ ions formed = Moles of NH4Cl - Moles of HClO4
Moles of NH4+ ions formed = 0.016 mol - 0.020 mol = -0.004 mol
Step 5: Calculate the final concentration of NH3
Final concentration of NH3 = Moles of NH3 / Final volume (L)
Final volume = Initial volume + Volume of HClO4 added
Final volume = 0.160 L + 0.020 L = 0.180 L
Final concentration of NH3 = -0.004 mol / 0.180 L = -0.022 M
Step 6: Calculate the final concentration of NH4+
Final concentration of NH4+ = Moles of NH4Cl + Moles of HClO4 / Final volume (L)
Final concentration of NH4+ = (0.016 mol + 0.020 mol) / 0.180 L = 0.178 M
Step 7: Calculate the Kb value of NH3
Kb = Kw / Ka
Since NH4+ is the conjugate acid of NH3, we can use the Kb of NH4+
Ka of NH4+ = Kw / Kb
Ka of NH4+ = 1.00 × 10^-14 / 1.77 × 10^-5 = 5.65 × 10^-10
Step 8: Calculate the concentration of OH- ions
OH- ions concentration = sqrt(Kw / Ka) × NH4+ concentration
OH- ions concentration = sqrt(1.00 × 10^-14 / 5.65 × 10^-10) × 0.178 M = 0.00662 M
Step 9: Calculate the pOH
pOH = -log10(OH- ions concentration) = -log10(0.00662) = 2.18
Step 10: Calculate the pH
pH = 14 - pOH = 14 - 2.18 = 11.82
Therefore, the pH of the new solution after adding 20.00 mL of 1.00 M HClO4 is 11.82.
Now, let's move on to part (b) and calculate the pH after adding another 20.00 mL of the same acid solution.
Step 1: Calculate the moles of excess HClO4
Moles of HClO4 = Volume (L) × Concentration (mol/L)
Moles of HClO4 = 0.020 L × 1.00 mol/L = 0.020 mol
Step 2: Determine the moles of NH4+ ions formed
Moles of NH4+ ions formed = Moles of HClO4
Moles of NH4+ ions formed = 0.020 mol
Step 3: Calculate the final concentration of NH4+
Final concentration of NH4+ = (Initial concentration × Initial volume + Moles of NH4+ ions formed) / Final volume (L)
Final volume = Initial volume + Volume of HClO4 added
Final volume = 0.160 L + 0.040 L = 0.200 L
Final concentration of NH4+ = (0.100 mol/L × 0.160 L + 0.020 mol) / 0.200 L = 0.088 M
Step 4: Calculate the Kb value of NH4+
Kb = Kw / Ka
Since NH4+ is the conjugate acid of NH3, we can use the Kb of NH4+
Ka of NH4+ = Kw / Kb
Ka of NH4+ = 1.00 × 10^-14 / 1.77 × 10^-5 = 5.65 × 10^-10
Step 5: Calculate the concentration of OH- ions
OH- ions concentration = sqrt(Kw / Ka) × NH4+ concentration
OH- ions concentration = sqrt(1.00 × 10^-14 / 5.65 × 10^-10) × 0.088 M = 0.00448 M
Step 6: Calculate the pOH
pOH = -log10(OH- ions concentration) = -log10(0.00448) = 2.35
Step 7: Calculate the pH
pH = 14 - pOH = 14 - 2.35 = 11.65
Therefore, the pH of the solution after adding another 20.00 mL of the same acid solution is 11.65.