How would you make 100 mL of a carbonic acid buffer at 0.5 M and pH = 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl and water?
so far, I have 50 mL NaHCO3. I plugged that into the Henderson-Hasselbalch equation of pH=pKa + log [A-]/HA] and knowing that [Buffer]= [A-] + [HA] , and get [HA] = 0.5 M. Meaning that [A-] is zero... And that doesn't sound right. I'm thinking I used the wrong pKa value? I have the equation for the dissociation of H2CO3 with the pKa being 6.35, and the equation of HCO3 dissociating with a pKa of 10.33. I would think I would use 10.33 since I am starting with NaCO3, but clearly I'm messing up somewhere. Anything would help me at this point!! Thanks in advance for your time.
You should use k1. You are starting with NaHCO3, not Na2CO3. And (base) + (acid) = 0.5M
For buffers you want to use a pKa within +/- 1 of the desired pH.
To make a carbonic acid buffer at 0.5 M and pH 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl, you need to use the correct pKa value for the carbonic acid system.
The Henderson-Hasselbalch equation for the carbonic acid system can be written as:
pH = pKa + log([HCO3-]/[H2CO3])
The pKa values you mentioned are correct: 6.35 for H2CO3 and 10.33 for HCO3-. However, H2CO3 is not present in the buffer system as it is a weak acid that rapidly dissociates into CO2 and H2O. Therefore, we will be using the pKa of HCO3-.
Step 1: Calculate the concentration of H2CO3 (which is essentially zero)
[H2CO3] = 1.0 M NaHCO3 * (100 mL / 1000 mL) = 0.1 M
Step 2: Calculate the concentration of HCO3-
[HCO3-] = 0.5 M - [H2CO3]
Since [H2CO3] is negligible, [HCO3-] ≈ 0.5 M.
Step 3: Calculate the concentration of Na+ (sodium ion)
[Na+] = [HCO3-]
So, the concentration of Na+ is also approximately 0.5 M.
Step 4: Determine the volume of NaHCO3 required.
0.5 M NaHCO3 * V = 0.5 M * 100 mL
V = 100 mL
So, you would need 100 mL of 1.0 M NaHCO3 for the buffer solution.
To achieve pH 6.0, you can either use 1.0 M NaOH or 1.0 M HCl. Here's how to calculate the volume required for each case:
Using 1.0 M NaOH:
pH = pKa + log([HCO3-]/[H2CO3])
6.0 = 10.33 + log([HCO3-]/[H2CO3])
log([HCO3-]/[H2CO3]) = -4.33
[HCO3-]/[H2CO3] = 10^-4.33
[HCO3-] ≈ 10^-4.33 * [H2CO3] ≈ 10^-4.33 * 0.1 M ≈ 5.01 × 10^-5 M
Since [HCO3-] ≈ 0.5 M, [H2CO3] ≈ 0.1 M, and NaOH is a strong base, we can assume that the concentration of OH- is equal to 0.5 M.
Using the equation: [OH-] * V = [H2CO3] * (100 mL / 1000 mL)
0.5 M * V = 0.1 M * (100 mL / 1000 mL)
V = 20 mL
Therefore, you would need to add 20 mL of 1.0 M NaOH to the 100 mL of 1.0 M NaHCO3 to make a carbonic acid buffer at pH 6.0.
Alternatively, if you choose to use 1.0 M HCl:
pH = pKa + log([HCO3-]/[H2CO3])
6.0 = 10.33 + log([HCO3-]/[H2CO3])
log([HCO3-]/[H2CO3]) = -4.33
[HCO3-]/[H2CO3] = 10^-4.33
[HCO3-] ≈ 10^-4.33 * [H2CO3] ≈ 10^-4.33 * 0.1 M ≈ 5.01 × 10^-5 M
Since [HCO3-] ≈ 0.5 M, [H2CO3] ≈ 0.1 M, and HCl is a strong acid, we can assume that the concentration of H+ is equal to 0.5 M.
Using the equation: [H+] * V = [HCO3-] * (100 mL / 1000 mL)
0.5 M * V = 0.5 M * (100 mL / 1000 mL)
V = 100 mL
Therefore, you would need to add 100 mL of 1.0 M HCl to the 100 mL of 1.0 M NaHCO3 to make a carbonic acid buffer at pH 6.0.
I hope this helps! Let me know if you have any further questions.
To make a carbonic acid buffer, you need to consider the dissociation of carbonic acid (H2CO3) into bicarbonate ion (HCO3-) and a proton (H+):
H2CO3 ⇌ H+ + HCO3-
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid (in this case, carbonic acid) and the ratio of the concentration of the conjugate base (in this case, bicarbonate ion) to the concentration of the acid:
pH = pKa + log([A-]/[HA])
In this case, you are given a 1.0 M NaHCO3 (bicarbonate) solution and need to dilute it to create a 0.5 M solution with a pH of 6.0. To determine the correct pKa value to use, we will examine the dissociation equations you mentioned.
The dissociation of H2CO3 has a pKa value of 6.35, and the dissociation of HCO3- has a pKa value of 10.33. To calculate the pH of a carbonic acid buffer, we need to use the pKa value for the dissociation of H2CO3.
Let's go through the calculations:
1. Determine the buffer concentrations:
- You have 50 mL of 1.0 M NaHCO3, which is 50 mmol NaHCO3.
- To prepare a 0.5 M solution, you need to dilute the NaHCO3 by a factor of 2. Therefore, you would need 25 mL of the 1.0 M NaHCO3.
2. Convert mL to L:
- 25 mL = 0.025 L
3. Calculate the number of moles (mmol) of NaHCO3:
- 0.025 L x 1.0 M = 0.025 mol NaHCO3 (25 mmol NaHCO3)
4. Calculate the number of moles of HCO3- in the solution:
- Since NaHCO3 dissociates to produce one mole of HCO3- for each mole of NaHCO3, you have 0.025 mol HCO3-
5. Calculate the number of moles of H2CO3 in the solution:
- Since H2CO3 initially does not exist, we can assume that all the H2CO3 is generated from the HCO3-. Therefore, you have 0.025 mol H2CO3.
6. Calculate the concentration of H2CO3:
- Divide the moles of H2CO3 (0.025 mol) by the total volume (0.100 L):
=> 0.025 mol / 0.100 L = 0.25 M H2CO3
So, your carbonic acid buffer will have a concentration of 0.25 M H2CO3 and a pH of 6.0. This pH is lower than the pKa, which is expected for a buffer with a ratio of [A-] to [HA] less than 1.
Note: The pKa value you mentioned for the dissociation of H2CO3 is slightly different from the commonly accepted value, which is around 6.35. It's crucial to use the correct pKa value to ensure accurate calculations.