Here are 2 questions I worked out, but need them checked. Also, can't get answer to another question correct unless question 1 is correct. Would you check the first 2 and help me with the third?
1) What is the pH of a 0.05 M solution of TRIS acid (pka = 8.3)?
My answer: I used the formula pH= (pka- log (HA)} /2 then I plugged in:
pH= 8.3-log(0.050)/2
pH= 4.8
2) What is the pH of a 0.045 M solution of TRIS base?
My answer: I used the formula PH=pka+14+log (base)/2
Then i plugged in:
pH=8.3+14+log(0.045M)/2 =
pH=10
3) How many total mL of 1 M NaOH can you add to the solution in problem 1 and still have a good buffer (that is, within 1 pH unit of the pka)?
Thank you.
Let's check your answers to the first two questions and then help you with the third one.
1) To find the pH of a solution of TRIS acid (pka = 8.3), you correctly used the Henderson-Hasselbalch equation. However, there is a mistake in your calculation. The correct formula is:
pH = pka + log([A-]/[HA])
In this case, TRIS acid (HA) acts as an acid and donates a proton (H+) to form the conjugate base, TRIS-, represented as (A-). Since the concentration of TRIS acid is given as 0.05 M, and assuming the concentration of TRIS- is negligible compared to the acid, the equation becomes:
pH = 8.3 + log(0.05)
pH ≈ 8.3 + (-1.3)
pH ≈ 6.98
Therefore, the pH of the 0.05 M TRIS acid solution is approximately 6.98.
2) To find the pH of a solution of TRIS base, you made a mistake in your formula. The correct Henderson-Hasselbalch equation for a base is:
pH = pka + log([base]/[acid])
In this case, TRIS base is considered the base, and its conjugate acid is TRIS acid. Assuming the concentration of TRIS base is 0.045 M and the concentration of TRIS acid is negligible compared to the base, the equation becomes:
pH = 8.3 + log(0.045/0.05)
pH ≈ 8.3 + log(0.9)
pH ≈ 8.3 + (-0.05)
pH ≈ 8.25
Therefore, the pH of the 0.045 M TRIS base solution is approximately 8.25.
Now, let's move on to the third question.
3) To determine how much 1 M NaOH can be added to the TRIS acid solution while maintaining a good buffer within 1 pH unit of the pKa, we need to consider the buffer capacity. The buffer capacity is highest when the concentrations of the acid and base involved in the buffer system are closest to each other.
In this case, you have a TRIS acid solution with a pH of approximately 6.98, which means it is already a weak acid buffer. To remain within 1 pH unit of the pKa (8.3), we need to consider the acceptable pH range, which would be approximately 7.3 to 9.3.
To calculate the amount of 1 M NaOH that can be added, we need to determine the moles of TRIS acid present in the initial solution and then add the same number of moles of NaOH.
Let's assume the initial volume of the TRIS acid solution is V mL.
No. of moles of TRIS acid in the initial solution = 0.05 mol/L × (V mL / 1000 mL)
Since we want to maintain the same moles of acid and base after adding NaOH:
No. of moles of NaOH to be added = 0.05 mol/L × (V mL / 1000 mL)
To convert moles to mL, we use the molarity of NaOH:
Volume of 1 M NaOH to be added = (0.05 mol/L × (V mL / 1000 mL)) × (1000 mL / 1 mol)
Simplifying the expression:
Volume of 1 M NaOH to be added = 0.05 V mL
So, the total mL of 1 M NaOH that can be added is 0.05 times the initial volume (V) of the TRIS acid solution.
Please provide the initial volume of the TRIS acid solution (V) to calculate the amount of NaOH that can be added.
To check your answers for the first two questions:
1) To determine the pH of a solution based on the pKa value, you need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, TRIS acid (HA) acts as a weak acid and dissociates into its conjugate base (A-). Since the pKa is given as 8.3, you can calculate the pH:
pH = 8.3 + log ([A-]/[HA])
pH = 8.3 + log (0.05/1)
Calculating this, the correct answer is pH = 7.3, not 4.8 as you mentioned.
2) For the second question, you are dealing with TRIS base, which is the conjugate base. It acts as a weak base and accepts a proton (H+). In this case, you need to consider the pKa value and calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Since TRIS base (A-) is the base, and the corresponding acid is water (HA), you can calculate the pH:
pH = 8.3 + log (0.045/1)
Calculating this, the correct answer is pH = 8.3 - 0.35 ≈ 7.95, not 10 as you mentioned.
Now, let's move on to the third question:
3) To determine how much 1 M NaOH you can add to the TRIS acid solution and still maintain an effective buffer, you need to consider the pH range within 1 pH unit of the pKa.
The pKa of TRIS acid is 8.3, so a pH range within 1 unit is 7.3 to 9.3.
To achieve this range, you need to convert the concentration of TRIS acid to moles:
moles of TRIS acid = volume (in L) x molarity
moles of TRIS acid = 0.05 M x volume (in L)
To maintain an effective buffer, the moles of the TRIS acid and the moles of the added NaOH should be equal or close to equal.
moles of TRIS acid = moles of NaOH
So, you can set up the following equation:
0.05 M x volume (in L) = 1 M x volume (in L)
Solving for the volume of NaOH in liters:
volume (in L) = (0.05 M x volume (in L)) / 1 M
Now, we can plug in the values:
volume (in L) = (0.05 x volume (in L)) / 1
1 = 0.05 x volume (in L)
volume (in L) = 1 / 0.05
volume (in L) = 20 L
Finally, convert the volume from liters to milliliters:
volume (in mL) = 20 L x 1000 mL/L
volume (in mL) = 20000 mL
Therefore, you can add a maximum of 20000 mL (or 20 liters) of 1 M NaOH to the solution in problem 1 and still have a good buffer within 1 pH unit of the pKa.