[NYM]

Calculate pH for

1) 3,5 moles of propanedioic acid and 6 moles of potassium hydroxide

2) 5 moles of propanedioic acid and 4 moles of potassium hydroxide

V is 1 L.

COOH-CH₂-COOH + OH^- -> COOH-CH₂-COO^- + H₂O

and

COOH-CH₂-COO^- + OH^- -> ^-OOC-CH₂-COO^- + H₂O


Solution:
1) Hmm... propanedioic acid is a dicarboxylic acid, so n(COOH-CH₂-COOH ) = 0 moles, n(COOH-CH₂-COO^-) = 3,5 mole and n(^-OCC-CH₂-COO^-) = 2,5 mole.
Right?
And then I just use the Henderson-Hasselbalch equation.

2) I just want to check if 1) is ok, no need to write something stupid twice

[FeLiXe]

Solution:
1) Hmm... propanedioic acid is a dicarboxylic acid, so n(COOH-CH₂-COOH ) = 0 moles, n(COOH-CH₂-COO^-) = 3,5 mole and n(^-OCC-CH₂-COO^-) = 2,5 mole.

which makes 6 mole in total

the monoanion gets used up, too and you only have 1 mole of it

besides that it looks good

[NYM]

the monoanion gets used up, too and you only have 1 mole of it

You'at talking about 2), right?
Or else I'm confuzzled

[FeLiXe]

when you start out with 3.5 moles propanedionic acid you cannot have 3.5 moles monoanion and 2.5 moles dianion. where does all that come from?

[NYM]

Err... If I mix 3,5 moles of COOHCH₂COOH and 6 moles of KOH, I get 3,5 moles of COOH-CH₂-COO^-  and  2,5 moles of ^-OCC-CH₂-COO^-), because OH^- is a strong base.

[AWK]

3.5 moles of hydrogenpropanedionate reacts with 2.5 mole of KOH to form 2.5 mole of potassium propanedionate (malonate), and  1.0 mole of  hydrogenpropanedionate remains unreacted!

[NYM]

Ahh, I see

Thanks