What Concentrations of KH2PO4 and Na2HPO4 are Needed to Make a pH 7.00 Buffer?

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In summary: For the KOH titration, the equivalence point will be the point at which the KOH solution has a pH equal to the acetic acid solution's pH.
  • #1
Gata
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can someone please help me with this problem, its giving me lots of trouble to solve :

(a) Suppose you wanted to make a buffer of exactly ph 7.00 using KH2PO4 and Na2HPO4. If you had a solution of 0.1M KH2PO4, what concentration of Na2HPO4 would you need?

(b)Now assume you wish to make a buffer at the same ph, using the same substances, but want the total phosphate molarity (HPO42-) + (H2PO4-) to equal 0.3 What concentrations of KH2PO4 and Na2HPO4 would you use?
 
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  • #2
What have you tried so far ? What ideas do you have ?

Tell us what you've tried and we'll guide you the rest of the way. :smile:
 
  • #3
i tried the henderson/hasselbalch equation :
the salt : Na2hpo4
the acid :kh2po4
ph= 7.0
pka ? not sure
not sure how to proceed on the second part or how to finish the first part
 
  • #4
If you want to find pH, then you should use pKa of the acid in Henderson-Hasselbach equation.
In your second question, you don't know the molar ratio of conjugate base to acid. The rest is known, just find it.
 
  • #5
question

its been said that the pka of strong bases are so close to 1 they ( the pka ) can be ignored.

Is this true ?
 
  • #6
Yes that's correct, but it is invalid for this question, as phosphate species other than ortho-phosphoric acid (H3PO4) need pKa values. If you were not given that data, it is nearly not possible to solve the H-H equation. Please find the pKa values for them, and solve this question. It is not that difficult. But when you think you're lost, keep asking.
 
  • #7
thanks

i have already solved the problem i was just asking a general question
thanks chem_tr
 
  • #8
new question

I was given this lab problem calc the ph values and draw the titration curve for the titration of 500ml of 0.010m acetic acid (pka 4.76) with .010m KOH.
I have to give two points the starting ph and the equivalence point.

I found the equivalence point to be the pka ( where ph=pka ).
I know i have to calc the second point using the KA I am just not sure how to plot the x-axis and y-axis

for the y-axis i know i have to plot ph
but for the x-axis is it KOH (eq ) or what ?
im not sure
can someone help me with this one ?
 
  • #9
Well, the starting pH will be the pH of the acetic acid solution with no KOH added, I am sure you are not asking this. What you're asking instead is simple, too; just plot the curve between titer (mL of KOH solution) and pH, on x and y axis, respectively.
 

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