CHEM 4 - Calculating ph from weak acid and strong base - HELP NEEDED Watch

danipertusiello
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For this question:

Calculate the pH of the solution formed when 50 cm3 of 0.500 mol dm-3 chloroethanoic acid (pKa = 2.86) is added to 25 cm3 of 0.100 mol dm-3 Ba(OH)2.

So far I have:

moles of chloroethanoic acid = 50x10-3 x 0.500 = 0.025 mol
moles of ba(oh)2 = 25x10-3 x 0.100 = 0.0025 mol

so the acid is in excess.

moles of chloroethanoic acid remaining = 0.025 - 0.0025 = 0.225 mol

However considering there is 2 OH molecules in Ba(OH)2 what must I do next in this question?
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charco
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(Original post by danipertusiello)
For this question:

Calculate the pH of the solution formed when 50 cm3 of 0.500 mol dm-3 chloroethanoic acid (pKa = 2.86) is added to 25 cm3 of 0.100 mol dm-3 Ba(OH)2.

So far I have:

moles of chloroethanoic acid = 50x10-3 x 0.500 = 0.025 mol
moles of ba(oh)2 = 25x10-3 x 0.100 = 0.0025 mol

so the acid is in excess.

moles of chloroethanoic acid remaining = 0.025 - 0.0025 = 0.225 mol

However considering there is 2 OH molecules in Ba(OH)2 what must I do next in this question?
If the chloroethanoic acid is in excess there is no barium hydroxide remaining in the final mixture!

You simply have to find out the moles of ethanoate ions produced from the reaction and then use the buffer relationship derived from:

ka = [H+][A-]/[HA]

as all components are in the same volume then just fill in the moles of HA and A- and rearrange to get [H+]
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danipertusiello
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(Original post by charco)
If the chloroethanoic acid is in excess there is no barium hydroxide remaining in the final mixture!

You simply have to find out the moles of ethanoate ions produced from the reaction and then use the buffer relationship derived from:

ka = [H+][A-]/[HA]

as all components are in the same volume then just fill in the moles of HA and A- and rearrange to get [H+]
Thanks for your reply!

So basically this is my equation:

HA + OH- = A- + H2O

HA before: 0.025
HA after: 0.0225

OH- before: 0.0025
OH- after: 0

A- before = 0
A- after = 0.0025

I then did....

10^-2.86 x (0.0225/75x10-3)

(0.0025/75x10-3)

= 0.0124

therefore ph = -log(0.0124) = 1.91

However the answer on the mark scheme is: 2.26

any ideas?
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danipertusiello
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Anyone??
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danipertusiello
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Hello
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charco
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(Original post by danipertusiello)
Thanks for your reply!

So basically this is my equation:

HA + OH- = A- + H2O


HA before: 0.025
HA after: 0.0225

OH- before: 0.0025
OH- after: 0

A- before = 0
A- after = 0.0025

I then did....

10^-2.86 x (0.0225/75x10-3)

(0.0025/75x10-3)

= 0.0124

therefore ph = -log(0.0124) = 1.91

However the answer on the mark scheme is: 2.26

any ideas?

There's your problem ...

The reaction between chloroethanoic acid and barium hydroxide is:

2ClCH2COOH + Ba(OH)2 ---> (ClCH2COO)2Ba + 2H2O
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username1445490
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Because the reacting ratio is 2:1 then to find the moles of HA after you have to subtract 2 x the moles of OH- that reacted whereas you subtracted the same number of moles as OH-.
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