1.What is the pH of a solution containing 0.042 M NaH2PO4 and 0.058 M Na2HPO4 ? The pKa of sodium phosphate is 6.86. pH = 6.86 + log [0.058] / [0.042]; pH = 6.86 + 0.14; pH = 7.00
That is my answer to the above question. The following question is what I am having trouble with.
2.What is the pH of the sodium phosphate buffer described in problem #1 if 1 ml of 10M NaOH is added? What is the pH of 1 liter of water if 1ml of NaOH is added?
2a. Adding 1 mL of 10M NaOH to HOW MUCH of the buffer. Since the problem gives molarities, I will assume 1L.
millimols HPO4^2- = 1000 mL x 0.058 = 58
mmols H2PO4 = 1000 mL x 0.042 = 42
mmols NaOH added = 1 mL x 10M = 10
.......H2PO4^- + OH^- ==> HPO4^2- + H2O
I......42........0........58.........0
added...........10.................
C......-10.....-10........+10.....+10
E.......32......0.........68
pH = 6.86 + log(68/32) = about 7.19
2b.
10M NaOH x (1/1001) = 0.00999M
pOH = 2
pH = 10
Check my work for typos. The chemistry is ok.
Well, well, well, looks like we have a pH expert in the house! But don't worry, I'll still try to put a smile on your face with my answer.
So, in problem #1, you found that the pH of the sodium phosphate solution is 7.00. Now, in problem #2, we have to add 1 ml of 10M NaOH to the solution. Let's see what happens.
When you add NaOH to a solution, it reacts with the acid in the buffer, which in this case is NaH2PO4. This reaction forms water and Na2HPO4.
Now, Na2HPO4 is a weak base, but we can ignore that for now. We just need to focus on the pH change due to the reaction between NaOH and NaH2PO4.
To calculate the new pH, we need to consider the amount of base added and the ratio of the acid and the conjugate base.
Hmm, did I hear someone say math? Don't worry, I'll make it fun! Imagine you're at a circus and you just threw a ball at a clown's face. The amount of laughter (pH change) depends on how hard you threw the ball (amount of base) and the clown's face elasticity (acid/base ratio).
So, in problem #2, with 1 ml of 10M NaOH added, the pH will change. But remember, the clown's face (buffer system) will try to resist that change, so the pH won't move too far.
Now, for the pH of 1 liter of water with 1 ml of NaOH added, well, that's a different story. Water is innocent and pH-neutral on its own, but add some NaOH to it, and it's like you've given the water a spicy surprise! It'll definitely change the pH, but I won't spoil the surprise for you. Just keep in mind that water can also act as a buffer, but it's not as good as sodium phosphate.
So, grab a popcorn and let's see how these pH changes play out in the circus of chemistry!
To calculate the new pH after adding 1 mL of 10 M NaOH to the sodium phosphate buffer described in problem #1, we need to consider the dissociation of NaOH and its effect on the buffer system.
Step 1: Calculate the moles of NaOH added.
Moles of NaOH = concentration of NaOH × volume of NaOH
Moles of NaOH = 10 M × 0.001 L
Moles of NaOH = 0.01 mol
Step 2: Determine the moles of sodium phosphate in the buffer.
In the problem description, we have the concentrations of the two components of the buffer solution, NaH2PO4 and Na2HPO4.
Moles of NaH2PO4 = concentration of NaH2PO4 × volume of the buffer solution
Moles of NaH2PO4 = 0.042 M × 1 L
Moles of NaH2PO4 = 0.042 mol
Moles of Na2HPO4 = concentration of Na2HPO4 × volume of the buffer solution
Moles Na2HPO4 = 0.058 M × 1 L
Moles of Na2HPO4 = 0.058 mol
Step 3: Calculate the new moles of the components in the buffer solution.
The addition of NaOH will react with the acidic component (NaH2PO4) of the buffer to form the conjugate base (HPO4^2-). The amount of NaH2PO4 consumed will be equal to the amount of NaOH added, and the amount of Na2HPO4 will remain unchanged.
Moles of NaH2PO4 remaining = moles of NaH2PO4 initial - moles of NaOH added
Moles of NaH2PO4 remaining = 0.042 mol - 0.01 mol
Moles of NaH2PO4 remaining = 0.032 mol
Step 4: Calculate the new concentrations of the components in the buffer solution.
Concentration of NaH2PO4 = Moles of NaH2PO4 remaining / volume of the buffer solution
Concentration of NaH2PO4 = 0.032 mol / 1 L
Concentration of NaH2PO4 = 0.032 M
Concentration of Na2HPO4 remains unchanged at 0.058 M.
Step 5: Calculate the new pH of the buffer.
pH = pKa + log [concentration of HPO4^2-] / [concentration of H2PO4^-]
pH = 6.86 + log [0.058] / [0.032]
pH ≈ 6.86 + 0.17
pH ≈ 7.03
Therefore, the pH of the sodium phosphate buffer after adding 1 mL of 10 M NaOH is approximately 7.03.
Regarding the pH of 1 liter of water after adding 1 mL of NaOH, water is neutral with a pH of 7. Adding a small amount of NaOH should not significantly affect the pH of the water.
To solve the first part of the question, you need to consider the reaction that occurs when NaOH is added to the sodium phosphate buffer. NaOH is a strong base that reacts with the acidic species (H2PO4-) in the buffer to form water and the conjugate base (HPO4^2-).
The balanced equation for the reaction is:
H2PO4- + OH- -> HPO4^2- + H2O
Since 1 mL of 10 M NaOH is added, you can calculate the moles of OH- added:
moles of OH- = volume (in L) * concentration (in M)
moles of OH- = 0.001 L * 10 M = 0.01 moles of OH-
Now the initial amount of H2PO4- was given as 0.042 M, but since some of it will react with OH-, the final concentration of H2PO4- will change. You can calculate the new concentration of H2PO4- by subtracting the moles of OH- reacted from the initial amount of H2PO4-.
moles of H2PO4- reacted = moles of OH- added
new moles of H2PO4- = initial moles of H2PO4- - moles of H2PO4- reacted
new moles of H2PO4- = 0.042 moles - 0.01 moles = 0.032 moles
To find the new concentration of H2PO4-, divide the new moles of H2PO4- by the volume in liters:
new concentration of H2PO4- = new moles of H2PO4- / volume (in L) = 0.032 moles / 0.001 L = 32 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log([HPO4^2-] / [H2PO4-])
pH = 6.86 + log(32/0.042)
pH = 6.86 + 1.3
pH = 8.16
So, the pH of the sodium phosphate buffer after adding 1 mL of 10M NaOH is 8.16.
Now for the second part of the question:
When 1 mL of NaOH is added to 1 liter of water, the concentration of OH- will be given by:
moles of OH- = volume (in L) * concentration (in M)
moles of OH- = 0.001 L * 10 M = 0.01 moles of OH-
Since there is only water, there is no acidic species to react with OH-, and hence no change in concentration of H+ (protons). Therefore, the pH of 1 liter of water is still 7 (neutral).
Therefore, after adding 1 mL of NaOH to 1 liter of water, the pH remains at 7.