Question

I need to calculate the amount of NaH2PO4 (in grams) and 1.00M NaOH solution (in ml) needed to make 250 ml of 0.20M H2PO4- at pH 8.00.

My work is lengthy and involved but the answer I got was 7.00 ml of the NaOH and 0.0070 moles of the NaH2PO4 needed.

Can you tell me if I am correct? If not I will retry the calculation.

Answer 1

If we are to try for the same number I need to know what you used for pK2.

Answer 2

7.21

Answer 3

I don't think so.
pH = 7.21 + log (0.007/0.007)
pH = 7.21 which isn't 8.00
7 mL x 1 M = 7 millmols NaOH = 0.007 mols.

Answer 4

Hmmm, ok. True. Ok I will try again. Thank you.

Answer 5

Can you suggest a different pka or what I might have done wrong? I got a pka of 7.2 by using pKa = -log(Ka) and Ka for H2PO4- is 6.3e10^-8. I know that a pKa of 7 will yield a buffer in the range of 6.2 to 8.2.

I calculated I needed 6.1 g of the NaH2PO4 and 7 ml of the 1.00M naOH to bring to a volume of 250 ml by titration to make a pH 8.

I got 6.1 g of NaH2PO4... gads!

Answer 6

7.2 is right for pK2.
And I obtained 6.0g NaH2PO4 which is quite close to your number.
The difference is that I used 43.15 mL of 1M NaOH to do that and that's the difference.
Do you want to work on that a little or want me to show you what I did.

Answer 7

Could you show me how you got it? I'm not sure what I did wrong.