PartA: What volume of 10.0M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? TrisHCl is a weak base and the molecular weight is 157.67 g/mol. I found the solution, it is 6.67 milliliters.
Part B: "The buffer from Part A is diluted to 1.00 L. To half of it (500.mL), you add 0.0150 mol of hydrogen ions without changing the volume. What is the resulting pH? pKb for the base= 5.91"
Part C
What additional volume of 10.0 would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?
Express your answer in milliliters using two significant figures.
JUST RESPONSE TO PART C ONLY!
thanks!
To calculate the additional volume of 10.0M NaOH needed to exhaust the remaining capacity of the buffer after the reaction described in Part B, we will use the Henderson-Hasselbalch equation:
pOH = pKb + log ([base]/[acid])
In this case, the base is TrisHCl and the acid is the undissociated form of Tris. Since the volume remains constant, the moles of hydrogen ions added (0.0150 mol) will react with an equal moles of the undissociated form of Tris.
First, let's calculate the initial concentrations of the base and acid before the reaction:
Base concentration ([base]) = mass of TrisHCl / molar mass of TrisHCl
= 31.52 g / 157.67 g/mol
= 0.200 mol
Acid concentration ([acid]) = [base] (since Tris is a weak base and completely dissociates)
= 0.200 mol
Now, let's calculate the concentration of the base and acid after the reaction:
Moles of base remaining = initial moles of base - moles of hydrogen ions added
= 0.200 mol - 0.0150 mol
= 0.185 mol
Moles of acid remaining = initial moles of acid (since it reacts completely)
= 0.200 mol
Now, let's calculate the pOH of the buffer solution after the reaction:
pOH = pKb + log ([base]/[acid])
= 5.91 + log (0.185/0.200)
Using a calculator, you should find that pOH = 5.879.
Now, let's calculate the pH of the buffer solution:
pH + pOH = 14
pH = 14 - pOH
= 14 - 5.879
Using a calculator, you should find that pH = 8.121.
Since the pOH and pH values only differ by 0.001, we can assume that the solution is still considered a buffer.
Now, let's move on to Part C:
To exhaust the remaining capacity of the buffer, we need to convert the remaining moles of acid (0.185 mol) to moles of base. Since the molar ratio between acid and base is 1:1, the moles of base remaining will also be 0.185 mol.
Using the formula:
Volume of 10.0M NaOH needed = moles of base remaining / molarity of NaOH
= 0.185 mol / (10.0 mol/L)
Calculating this equation, you should find that the volume of 10.0M NaOH needed is 0.0185 L, or 18.5 mL (rounded to two significant figures).
Therefore, the additional volume of 10.0M NaOH needed to exhaust the remaining capacity of the buffer is 18.5 mL.
To find the additional volume of 10.0M NaOH needed to exhaust the remaining capacity of the buffer after the reaction described in Part B, we need to calculate the remaining moles of TrisHCl and use it to determine the required volume of NaOH.
First, let's calculate the moles of TrisHCl initially used in Part A:
Given mass of TrisHCl = 31.52 g
Molecular weight of TrisHCl = 157.67 g/mol
Moles of TrisHCl = (mass of TrisHCl) / (molecular weight of TrisHCl)
= 31.52 g / 157.67 g/mol
= 0.1999 mol (approximately 0.200 mol)
In Part B, we added 0.0150 mol of hydrogen ions to the buffer, which reacts with an equal amount of the weak base TrisHCl. Therefore, the remaining moles of TrisHCl in the buffer after the reaction are:
Remaining moles of TrisHCl = Initial moles of TrisHCl - Moles of hydrogen ions added
= 0.200 mol - 0.0150 mol
= 0.185 mol
To calculate the additional volume of 10.0M NaOH needed, we can use the equation:
Moles of NaOH = Moles of TrisHCl
Now, we can rearrange the equation to solve for the required volume of NaOH:
Volume of NaOH = (Moles of TrisHCl) / (Molarity of NaOH)
Given Molarity of NaOH = 10.0 M,
Mothers of TrisHCl = 0.185 mol,
Volume of NaOH = 0.185 mol / 10.0 M
= 0.0185 L
= 18.5 mL (approximately)
Therefore, the additional volume of 10.0M NaOH needed to exhaust the remaining capacity of the buffer is approximately 18.5 mL (rounded to two significant figures).