[shakey1]

What volume of 0.2M NaOH is required to prepare 200ml of a 0.2M buffer solution of pH 4 (pKa of acid is 3.5)?


Ive started off using ph=pka+log(b/a) then im not sure what to do..

Thanks.

[mpatel90]

(Original post by shakey1)
What volume of 0.2M NaOH is required to prepare 200ml of a 0.2M buffer solution of pH 4 (pKa of acid is 3.5)?


Ive started off using ph=pka+log(b/a) then im not sure what to do..

Thanks.

LOL i am stuck on this exact same question....

[Ari Ben Canaan]

(Original post by mpatel90)
LOL i am stuck on this exact same question....

Could you link me to the full question ?

[mpatel90]

hey think I have managed to figure it out - but please double check...

Q) What Volume of NaOH is required to prepare 200ml of a 0.2M buffer with a pH of 4 and pKa of acid = 3.5?

Step 1: work out ratio of acid and base in buffer using Henderson Hasselbach Eqn

[H+] = Ka (a/b)

H+ = antilog pH = 1x10-4
Ka = antilog pKa = 3.16 x 10-4

putting that into above equation you get a/b = 0.3162

Step 2: Work out concentrations of acid and base in buffer

the total concentration of buffer is 0.2M therefore a+b= 0.2M
so you can set up simultaneous equations to work out a and b

a=0.2 - b and a= 0.3162b

solving these gives you a = 0.0480 M and b = 0.1519 M

Step 3: Working out volume of base required

You know that the total volume of buffer is 200ml

therefore the amount of b is in 200ml
i.e. 0.1519 molar in 200ml
so what volume of 200ml would have 0.2M in?

(0.1519/0.2) X 200ml = 151.9 ml

therefore the volume of NaOH reqd to make this buffer is 151.9ml

[Ari Ben Canaan]

(Original post by mpatel90)
x

Yes, that is correct.

Which exam board is this ?

[mpatel90]

(Original post by Ari Ben Canaan)
Yes, that is correct.

Which exam board is this ?

Thanks - umm the question isnt from A Level stuff, I got it from a set of questions given to us by my lecturer at Uni (Ist year pharmacy...)

[Ari Ben Canaan]

(Original post by mpatel90)
Thanks - umm the question isnt from A Level stuff, I got it from a set of questions given to us by my lecturer at Uni (Ist year pharmacy...)

OHHH !! No wonder this seemed a little different from the questions I'm used to

[arvin_infinity]

(Original post by mpatel90)
hey think I have managed to figure it out - but please double check...

Step 3: Working out volume of base required

You know that the total volume of buffer is 200ml

therefore the amount of b is in 200ml
i.e. 0.1519 molar in 200ml
so what volume of 200ml would have 0.2M in?

(0.1519/0.2) X 200ml = 151.9 ml

therefore the volume of NaOH reqd to make this buffer is 151.9ml

It would be appreciated if you explain the last part a bit more