A buffer is made up of .300 L each of .500 M KH2PO4 and .317 M K2HPO4. Assuming tha the volumes are additive, calculate
a) the pH of the buffer
b) the pH of the bugger after the addition of .0500 mol of HCl to .600 L of the buffer.
c) the pH of the buffer after the addition of .0500 mol of NaOH to .600 L of the buffer.
I know that [H+] is very nearly equal to Ka, and that pKa is very nearly equal to pH, so I figured that if I could find [H+] I could find pH for at least part a. My problem right now is that I'm having trouble writing out equations for this reaction. Do I treat each part as its own separate entity, so that I have
KH2PO4 --> K+ + H2PO4-
K2HPO4 --> 2K+ + HPO4-2
? Or do I have a reaction directly between KH2PO4 and K2HPO4, or neither? Also, how does the addition of HCl and NaOH affect the pH calculationwise? I figured that upon the addition of HCl and NaOH, water and neutral NaCl would be formed, which it seems like it would keep the pH constant after the addition of NaOH. But I'm still a little confused on this- if anyone could clarify this for me, that would be great.
a) With a mixture of KH2PO4 and K2HPO4, you are between the 1st and 2nd equivalence points in the titration of H3PO4 with KOH. To calculate the pH, use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
H3PO4 has k1, k2, and k3. For this problem, pKa is pK2. That makes the HPO4^-2 the base and H2PO4^- the acid. Straight forward HH equation.
b) When x mols HCl are added, x mols HCl reacts with x mols of the base which add x mols to the acid and subtracts x mols from the base. Plug those numbers into a new HH equation, using the same pK2 and calculate the new pH.
c) When x mols NaOH are added, x mols NaOH reacts with x mols of the acid and that adds x mols to the base and subtracts x mols from the acid. Plug those new numbers into the HH equation and recalculate pH using the same pK2.
Post your work if you get stuck.
To begin, let's first calculate the pH of the buffer (part a) using the Henderson-Hasselbalch equation:
pH = pKa + log (base)/(acid)
In this case, the acid is H2PO4^-, and the base is HPO4^2-. The pKa value we need to use is the pK2 value, which corresponds to the dissociation of the second H+ from phosphoric acid.
1) Determine the pK2 value:
Since phosphoric acid (H3PO4) has three ionizable hydrogens, we need to find pK2 specifically. The pK2 value for phosphoric acid is approximately 7.2.
2) Calculate the molar concentrations:
The volume of the buffer is 0.300 L. To calculate the moles of the acid and the base, we use the molarity of each component and multiply it by the volume of the buffer.
Moles of acid (H2PO4^-) = (0.500 M) * (0.300 L) = 0.150 moles
Moles of base (HPO4^2-) = (0.317 M) * (0.300 L) = 0.0951 moles
3) Calculate the ratio of base/acid:
ratio = (moles of base)/(moles of acid)
ratio = 0.0951/0.150 = 0.634
4) Apply the Henderson-Hasselbalch equation:
pH = 7.2 + log (0.634)
pH ≈ 6.7
Therefore, the pH of the buffer is approximately 6.7.
Now, let's move on to part b and part c:
b) When HCl is added to the buffer, it reacts with the base (HPO4^2-) and forms more acid (H2PO4^-). The moles of HCl added to the buffer will be equal to the moles of base consumed and the moles of acid formed.
In this case, 0.0500 mol of HCl is added to 0.600 L of the buffer. This means that the moles of acid will increase by 0.0500 mol and the moles of base will decrease by the same amount. Adjust the moles of acid and base accordingly and recalculate the ratio using the new values.
Once you get the new ratio, plug it into the Henderson-Hasselbalch equation with the pK2 value to calculate the new pH.
c) When NaOH is added to the buffer, it reacts with the acid (H2PO4^-) and forms more base (HPO4^2-). Similar to part b, the moles of NaOH added will be equal to the moles of acid consumed and the moles of base formed.
In this case, 0.0500 mol of NaOH is added to 0.600 L of the buffer. Adjust the moles of acid and base accordingly and recalculate the ratio using the new values.
Once you get the new ratio, plug it into the Henderson-Hasselbalch equation with the pK2 value to calculate the new pH.
Remember, the Henderson-Hasselbalch equation can be used in cases where the concentration of the acid and base is known, and the pKa value is also known. You can use this equation to calculate the pH of a buffer before and after the addition of acids or bases.