you have a 0.200L of solution containing 250mM phosphate buffer, pH 6.8.
An equivalent amount of HCl/NaOH with respect to the total amount of phosphate in the buffer --> mmol of NaH2PO3 + mmol of NaH2PO4 = 50 mmoles.
What would the pH of the above buffer solution be after you add the following with respect to the total amount of phosphate in the buffer:
1) 1/8 equivalent amount of NaOH
2) 1/8 equivalent amount of HCl
1) 5.89
2) 11.47
How did you solve it? I wasn't really looking for an answer but the process because I don't really get it.
To solve this problem, we need to understand the concept of acid-base equilibria and how the addition of acid or base affects the pH of a solution.
Let's break down the problem step by step:
Step 1: Calculate the initial concentration of phosphate ions (PO4^3-) in the solution.
Given:
Volume of solution = 0.200 L
Concentration of phosphate buffer = 250 mM
To convert mM to moles per liter (M), divide by 1000:
Concentration of phosphate buffer = 250 mM = 0.250 M
Now, multiply the concentration by the volume to get the initial number of moles of phosphate ions:
Initial moles of phosphate ions = 0.250 M * 0.200 L = 0.050 mol
Step 2: Determine the initial pH of the buffer solution.
Given:
pH = 6.8
Step 3: Calculate the number of moles of NaH2PO3 and NaH2PO4 that make up 50 mmol of phosphate ions.
Given:
Total moles of phosphate ions = 50 mmol
Since NaH2PO3 and NaH2PO4 have a 1:1 ratio, let's assume x moles of NaH2PO3 and (50 - x) moles of NaH2PO4. Therefore:
x + (50 - x) = 50 mmol
Now, we can solve for x:
x = 25 mmol
So, there are 25 mmol of NaH2PO3 and 25 mmol of NaH2PO4 in the buffer solution.
Step 4: Calculate the moles of NaOH to be added.
Given:
1/8 equivalent amount of NaOH
To calculate the 1/8 equivalent, divide the total amount by 8:
1/8 equivalent amount = (1/8) * 50 mmol = 6.25 mmol
Step 5: Calculate the moles of HCl to be added.
Given:
1/8 equivalent amount of HCl
Again, calculate the 1/8 equivalent:
1/8 equivalent amount = (1/8) * 50 mmol = 6.25 mmol
Step 6: Calculate the new total moles of phosphate ions after adding NaOH.
To get the new total moles of phosphate ions, we need to subtract the moles of NaOH added from the moles of NaH2PO3 initially present:
New total moles of phosphate ions = 25 mmol - 6.25 mmol = 18.75 mmol
Step 7: Calculate the new total moles of phosphate ions after adding HCl.
Similarly, we subtract the moles of HCl added from the moles of NaH2PO4 initially present to get the new total moles of phosphate ions:
New total moles of phosphate ions = 25 mmol - 6.25 mmol = 18.75 mmol
Step 8: Calculate the new concentration of phosphate ions.
First, calculate the new total moles of phosphate ions per liter:
New concentration of phosphate ions = (18.75 mmol) / 0.200 L = 93.75 mM
Step 9: Calculate the new pH of the buffer solution after adding NaOH.
To determine the pH, we need to know the pKa values of the phosphate buffer system. The pKa values correspond to the acid dissociation constants for the acidic and basic forms of phosphate.
In this case, the pKa values are:
pKa1 = 2.12
pKa2 = 7.21
pKa3 = 12.32
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
For the basic form, [A-] = 18.75 mmol
For the acidic form, [HA] = 6.25 mmol
Let's calculate the new pH:
pH = 7.21 + log(18.75 mmol / 6.25 mmol) = 7.516
Therefore, the pH of the buffer solution after adding 1/8 equivalent amount of NaOH is approximately 7.516.
Step 10: Calculate the new pH of the buffer solution after adding HCl.
Now, let's repeat the same steps to calculate the new pH after adding 1/8 equivalent amount of HCl.
The only difference is that for this step, we will use the pKa1 value instead of pKa2.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
For the basic form, [A-] = 6.25 mmol
For the acidic form, [HA] = 18.75 mmol
Calculating the new pH:
pH = 2.12 + log(6.25 mmol / 18.75 mmol) = 1.619
Therefore, the pH of the buffer solution after adding 1/8 equivalent amount of HCl is approximately 1.619.
Please note that these calculations assume ideal behavior, and the actual pH may vary due to other factors.