[uzi4u2]

hello again !

he mixing of which solutions results in a phosphate buffer with p.h=7.2 ?
1.10ml 1 M H3PO4 + 15 ml 1 M NaOH
2.10ml 1 M H3PO4 + 10 ml 1 M NaOH
3.10ml 1 M H3PO4 + 5ml    1 M NaOH
4.10ml 1 M H3PO4 + 25ml  1 M BAOH
 
how do i get this one ? i mean what the equation i should use ?  i hae no PKa :/

[Albert]

H₃PO₄

first ionization kA = 6.92 E-3
second ionizaton kA = 6.17 E-8
third ionization kA = 2.09 E-12

 

[sdekivit]

hint from previous post: pKa = 7.21 

[uzi4u2]

yeah , but how do i get the Pka ??  they dont give it to me , i mean the question is as i wrote it in the exam :/    what am i supposed to know it by heart ?   
isnt there a diffrent way of calculating this withought knowing the pka ?   

thx again

[Borek]

isnt there a diffrent way of calculating this withought knowing the pka ?   

No.

But question is not that hard - all you have to remember to be able to solve it is that H₂PO₄^-/HPO₄^2- buffer is used for preparation of buffers in the pH 7.0 range. Assuming one of the answers is right, everything else is logic

[uzi4u2]

okay in theory the first answer is the correct one , but i just dont get it !!
whats the equation ?     i mean we use the Henderson-Hasselbalch equation right ?
so we already have the ph which is 7.2 but i dont have the PKa  so i am still missing a factor ?!?!?

pH=pKa+log([A-]/[HA])     how in the name of our holy lord do i do it ?   

[sdekivit]

because pH ~ pK_a it directly follows that in the equilibrium

H₂PO₄^- + H₂O <--> HPO₄^2- + H₃O⁺

The ratio [HPO₄^2-]/[H₂PO₄^-] ~ 1 and thus [HPO₄^2-]~[H₂PO₄^-] (Hendersson-Hasselbalch)

And thus we first need 10 ml NaOH to dissociate H₃PO₄ completely to H₂PO₄^- and then another 5 mL NaOH to dissociate half of the H₂PO₄^- to HPO₄^2-

--> that's why a is the correct answer.

[uzi4u2]

i dont get it :/
i just dotn get it .
whats the calculation ?    show me a formula some numbers    pleaseeeeeeeee
these buffers are killing me softly i hate them and i hate their questions 

[Borek]

First of all - you have already wrote Henderson-Hasselbalch equation:

pH=pK_a+log([A^-]/[HA])

Note, that when [A^-]=[HA] pH=pK_a. So if you know that H₂PO₄^-/HPO₄^2- is used for buffers with pH close to 7, you have to look through the answers.

3.10ml 1 M H3PO4 + 5ml    1 M NaOH
2.10ml 1 M H3PO4 + 10 ml 1 M NaOH
1.10ml 1 M H3PO4 + 15 ml 1 M NaOH
4.10ml 1 M H3PO4 + 25ml  1 M BAOH

(I have sorted them slightly different then in original question).

Solution with 5mL of base contains equimolar buffer made of H₃PO₄ and H₂PO₄^-.
Solution with 10mL of base contains only H₂PO₄^- - no buffer here.
Solution with 15mL of base contains equimolar buffer made of H₂PO₄^- and HPO₄^2-.
Solution with 25mL of base contains equimolar buffer made of HPO₄^2- and PO₄^3-.

Of those four solutions only one can have pH close to 7.