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Chemistry buffers question help

A student adds 50 cm3 of 0.25 moldm^-3 butanoic axid to 50cm of 0.05 moldm-3 NaOH.... can someone explain to me why the conc of the acid is 0.1? I'm getting 0.125
Reply 1
Original post by ineedA
A student adds 50 cm3 of 0.25 moldm^-3 butanoic axid to 50cm of 0.05 moldm-3 NaOH.... can someone explain to me why the conc of the acid is 0.1? I'm getting 0.125


The question is, how did you get 0.125.
Equal volumes of both solutions, so 0.25 - 0.05 = 0.2, but the volume doubles, so 0.2 / 2 = 0.1
Reply 2
Original post by GPiph
The question is, how did you get 0.125.
Equal volumes of both solutions, so 0.25 - 0.05 = 0.2, but the volume doubles, so 0.2 / 2 = 0.1


I did (0.25X50)/1000 to get 0.0125 then to get the concentration (0.0125x1000)/100 to give 0.125... can you explain why this isn't right 😒
Reply 3
Original post by GPiph
The question is, how did you get 0.125.
Equal volumes of both solutions, so 0.25 - 0.05 = 0.2, but the volume doubles, so 0.2 / 2 = 0.1
and why with equal volumes can you just minibus the conc of Naoh off the conc of the acid.. I understand that when you double the volume to concentration will have
Reply 4
Original post by ineedA
and why with equal volumes can you just minibus the conc of Naoh off the conc of the acid.. I understand that when you double the volume to concentration will have


Do you realise there's a nuetralisation reaction happening, water is created
Reply 5
Original post by GPiph
Do you realise there's a nuetralisation reaction happening, water is created


Yes, but I don't understand why you can minus the concentration of the alkali off the concentration of the weak acid
Reply 6
Original post by ineedA
Yes, but I don't understand why you can minus the concentration of the alkali off the concentration of the weak acid


Explain why you think you can't.

The acid will dissociate as H+ is removed from solution, the neautralisation will go til one of reactants run out
Reply 7
Original post by GPiph
Explain why you think you can't.

The acid will dissociate as H+ is removed from solution, the neautralisation will go til one of reactants run out


Ok I partially understand, but if oH- ions are used in the neutralisation wouldn't this decrease the concentration of the base too?
Reply 8
Original post by ineedA
Ok I partially understand, but if oH- ions are used in the neutralisation wouldn't this decrease the concentration of the base too?


Yes, base will react completely. you will have the salt sodium butanoate in solution though.

anyways im not a chemist, hopefully someone will correct me if im wrong
Reply 9
Original post by GPiph
Yes, base will react completely. you will have the salt sodium butanoate in solution though.

anyways im not a chemist, hopefully someone will correct me if im wrong


mark scheme says concentration of alkali is 0.025 so I'm not sure🙃, thank you anyway
Reply 10
Original post by ineedA
mark scheme says concentration of alkali is 0.025 so I'm not sure🙃, thank you anyway


the concentration of sodium butanoate is 0.025, just divide 0.05 by 2 as volume doubles.
Have you written out the equation for the reaction?
Reply 11
Original post by GPiph
the concentration of sodium butanoate is 0.025, just divide 0.05 by 2 as volume doubles.
Have you written out the equation for the reaction?


I understand the equations , I just don't understand why my method I used earlier is wrong /:
Reply 12
Original post by ineedA
I understand the equations , I just don't understand why my method I used earlier is wrong /:


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Reply 13
Original post by GPiph
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Thank you so much, realised my mistake

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