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Author Topic: Calculate the pH of a buffer after adding NaOH  (Read 13763 times)

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sammyjo06

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Calculate the pH of a buffer after adding NaOH
« on: August 30, 2009, 11:58:57 AM »

This was a 2 part problem, and I believe I did the first part correctly; I just need help with the second.

The first part of the problem said this: A 0.25 M pH 8.5 phosphate buffer needs to be made. Give the amounts in grams of the acid and of the base you will use to make 1.5 L of this buffer.

I calculated that 22.67 g of K3PO4 and 46.72 g of K2HPO4 were needed. If someone finds this to be incorrect, I can give you my calculations to see where I went wrong. Otherwise, I'll assume these numbers are right.

The second part is where I am having trouble.

I need to calculate the pH of the resultant solution after adding 1.0 mL of a 1.0 M solution of NaOH to 100 mL of the buffer above.

I really had no clue where to begin, except that I remembered pH = -log [H+]

So (I don't know if this is right) I tried to calculate the [H+] for the solution before anything was added. The pH is 8.5, so 8.5 = -log [H+]

-8.5 = log [H+]. [H+] = 3.16 x 10^-9

NaOH I believe is [H+] = 10^-14

So then (here is where I will have a hard time explaining what I was trying to do, because I really don't know),  for NaOH:
1.0   M / 101 mL = .0099 M
For the buffer:
0.25M / 101 mL = .002475 M
So for the [H+] of the solution:
(3.162 x 10^-9 / .002475) + (10^-14 / .0099) = 1.2776 x 10^-6

pH = -log(1.2776 x 10^-6)

pH = 5.894
Obviously, that’s wrong. The pH would not have decreased on adding a base. But I have no clue where I went wrong, or if I’m even using the right equation. Any help would be much appreciated!
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Fridushka

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Re: Calculate the pH of a buffer after adding NaOH
« Reply #1 on: August 30, 2009, 12:08:13 PM »

Also i don't know where you went wrong, but why
Quote
NaOH I believe is [H+] = 10^-14
!
here you have to calulate the [-OH] because it's basic, the through kw find the [H+]
here what you wrote for the NaOH was the kw, by that you mean the strong base is nill! Maybe that's your mistake
allthough Im not sure.
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sammyjo06

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Re: Calculate the pH of a buffer after adding NaOH
« Reply #2 on: August 30, 2009, 05:48:17 PM »

Also i don't know where you went wrong, but why
Quote
NaOH I believe is [H+] = 10^-14
!
here you have to calulate the [-OH] because it's basic, the through kw find the [H+]
here what you wrote for the NaOH was the kw, by that you mean the strong base is nill! Maybe that's your mistake
allthough Im not sure.

Maybe I have this completely wrong, but I think the pH of 1.0 M NaOH is 14. If you calculate the pOH, wouldn't it be -log[NaOH] = -log(1.0M) = 0? Then 14 - 0 = 14. So the [H+], to put it in the same units as the rest of the problem, would be 14 = -log [H+], [H+] = 1.0 x 10^-14. Correct me if I am wrong about that, but that was my line of thought there.

Edit: I think I figured out the problem. I used mL, but when I converted to liters and solved the problem, I got a pH of 8.89, which makes a lot more sense.
« Last Edit: August 30, 2009, 06:01:51 PM by sammyjo06 »
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Borek

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Re: Calculate the pH of a buffer after adding NaOH
« Reply #3 on: August 30, 2009, 09:09:20 PM »

I calculated that 22.67 g of K3PO4 and 46.72 g of K2HPO4 were needed. If someone finds this to be incorrect, I can give you my calculations to see where I went wrong.

Please do. Mixture of these salts will have pH in the pH 11 range.
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sammyjo06

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Re: Calculate the pH of a buffer after adding NaOH
« Reply #4 on: September 01, 2009, 11:46:08 AM »

^Here's what I did.

pH = pka + log([A-]/[HA])

8.5 = 8.9 (I got this number from the handout we received in class) + log ([K3PO4]/[K2HPO4])

-.4 = log ([K3PO4]/[K2HPO4])
 [K3PO4]/[K2HPO4] = .39812
(.39812)(K2HPO4) = K3PO4

M = (mol/L)
.25 = (mol/1.5L)
mol = .375 mol solution

.375 mol = mol K3PO4 + mol K2HPO4
.375 mol - K2HPO4 = K3PO4

I then combined my two equations,

0.375 - K2HPO4 = K3PO4
-[0 - (.39812)(K2HPO4) = K3PO4]

K3PO4s cancel each other out, left with 0.375 - 1.39812(K2HPO4) = 0
K2HPO4 = .26822 mol
Therefore, K3PO4 = .10678 mol if I plug it into one of those equations.

For K3PO4, 212.27 g/mol x 0.10678 mol = 22.67g
For K2HPO4, 174.176 g/mol x 0.26822 mol = 46.72g


Where did I go wrong?


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