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

Please explain. I'm not really looking for an answer but how it works.
Thank you

for 1) I got 7.42 after trying. I wanted to confirm if it is right to see if I got the process right?

I have no idea what this question means. My best suggestion is that if you understand the question that you use the same kind of process I used for that phosphate buffer I did last night.

To determine the pH of the buffer solution after adding NaOH or HCl, we need to understand the concept of buffer capacity and the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])

Where:
pH = the pH of the buffer solution
pKa = the acid dissociation constant for the acid component of the buffer
[A-] = the concentration of the conjugate base
[HA] = the concentration of the acid

In this case, the buffer system is a mixture of NaH2PO3 (acid) and NaH2PO4 (conjugate base), with the acid/base pair being H2PO3(-)/H2PO4(-). The pKa for this acid/base pair can be determined from the dissociation constant (Ka) of H2PO4(-):

Ka = ([H+][HPO4^-])/[H2PO4^-]

Since pH 6.8 is provided, we can calculate the [H+] concentration using the equation:
[H+] = 10^(-pH)

Now, let's calculate the initial concentrations of [HA] and [A-] in the buffer solution:

Initial [HA] = 0.2L * 250 mM = 50 mmol
Initial [A-] = 0.2L * 250 mM = 50 mmol

1) Adding 1/8 equivalent amount of NaOH:
Since the equivalence is 50 mmol, 1/8 of that would be 50 mmol * 1/8 = 6.25 mmol NaOH.

Now, we need to distribute this 6.25 mmol between [HA] and [A-]. Since NaOH reacts with the acid (HA) to form water and the conjugate base (A-), we can subtract the amount reacted from [HA] and add it to [A-].

Final [HA] = Initial [HA] - 6.25 mmol
Final [A-] = Initial [A-] + 6.25 mmol

To find the new pH, we use the Henderson-Hasselbalch equation with these new values.

2) Adding 1/8 equivalent amount of HCl:
Again, the equivalence is 50 mmol, so 1/8 of that would be 50 mmol * 1/8 = 6.25 mmol HCl.

Now, we need to distribute this 6.25 mmol between [HA] and [A-]. Since HCl reacts with the base (A-) to form water and the acid (HA), and we can subtract the amount reacted from [A-] and add it to [HA].

Final [HA] = Initial [HA] + 6.25 mmol
Final [A-] = Initial [A-] - 6.25 mmol

Again, we can use the Henderson-Hasselbalch equation with these new values to calculate the new pH.

By considering the changes in concentration of the acid (HA) and conjugate base (A-) components of the buffer after adding the specified amount of NaOH or HCl, we can calculate the new pH using the Henderson-Hasselbalch equation.