A buffer solution of volume 100.0 mL is 0.150 M Na2HPO4(aq) and 0.100 M KH2PO4(aq). Refer to table 1.

(a) What are the pH and the pH change resulting from the addition of 80.0 mL of 0.0500 M NaOH(aq) to the buffer solution?
pH

pH change (include negative sign if appropriate)


(b) What are the pH and the pH change resulting from the addition of 10.0 mL of 1.0 M HNO3(aq) to the initial buffer solution?
pH

pH change (include negative sign if appropriate)

So I found the pH of the initial buffer solution to be 7.38 (using pka2=7.21). I am not sure how to set up this problem to add the NaOH to the buffer solution

for part a I tried using the HH equation, found moles of NaOH and used that for [base] and moles Na2HPO4 as [acid]...i used pKa 7.21...that was incorrect. For part b I used a similar method...plugging in HNO3 as the [acid] and Na2HPO4 moles as [base]

the other pka values for H3PO4 are:
Ka1 = 7.6E-3 ...... pKa1= 2.12
Ka2 = 6.2E-8 ...... pKa2= 7.21
Ka3 = 2.1E-13 .....pKa3= 12.68

I figured out the correct new pH for each respective part, but apparently the initial pH i found for the buffer solution is incorrect. I just need help in figuring that so I can calculate the change.

I initially thought the buffer pH was 7.38. Thanks!

NVM i was using the pKA instead of pH....i had it right afterall...guess it's just time for bed....

what's the answer then?

To determine the pH and pH change resulting from the addition of NaOH to the buffer solution, you need to consider the reaction between NaOH and the components of the buffer solution.

In this case, Na2HPO4 and KH2PO4 are acting as a conjugate acid-base pair, with Na2HPO4 being the base (accepting a proton) and KH2PO4 being its conjugate acid (donating a proton).

First, let's calculate the moles of Na2HPO4 and KH2PO4 in the initial buffer solution.

Moles of Na2HPO4 = (0.150 M) x (0.100 L) = 0.015 moles
Moles of KH2PO4 = (0.100 M) x (0.100 L) = 0.010 moles

Now, let's consider the addition of 80.0 mL of 0.0500 M NaOH. The reaction that occurs is:

Na2HPO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l)

Since Na2HPO4 is a base, it reacts with NaOH to form Na3PO4 and water. The moles of Na2HPO4 reacting with NaOH can be determined using the stoichiometry of the balanced equation.

Moles of Na2HPO4 reacting with NaOH = (0.0500 M) x (0.0800 L) = 0.004 moles

After the reaction, the remaining moles of Na2HPO4 = 0.015 moles - 0.004 moles = 0.011 moles

The moles of KH2PO4 remain unchanged.

Now, we need to calculate the concentrations of Na2HPO4 and KH2PO4 in the final solution after the addition of NaOH.

Final volume = Initial volume + Volume of NaOH added = (100.0 mL) + (80.0 mL) = 180.0 mL = 0.180 L

Concentration of Na2HPO4 in the final solution = (0.011 moles) / (0.180 L) = 0.0611 M
Concentration of KH2PO4 remains = 0.100 M

Now, using the Henderson-Hasselbalch equation, we can calculate the pH of the final solution:

pH = pKa + log ([base]/[acid])

For this buffer system, pKa = 7.21. Plugging in the values:

pH = 7.21 + log (0.0611/0.100) = 7.21 + (-0.157) = 7.05

So, the pH of the final buffer solution, after the addition of NaOH, is 7.05.

To calculate the pH change, you can subtract the pH of the final buffer solution from the pH of the initial buffer solution:

pH change = 7.05 - 7.38 = -0.33 (negative sign due to a decrease in pH)

Now, you can follow a similar approach to part (b) by considering the reaction between HNO3 and the components of the buffer solution.