[mehc]

I want to make buffer of 0.067 M potassium dihydrogen phosphate , of pH 7.5.

I am in problem because i don't have access to desired chemicals.

I have potassium dihydrogen phosphate. The methods i have found mostly use sodium phosphate dibasic, or potassium phosphate dibasic with potassium dihydrogen phosphate to form pH 7.5 buffer.

So kindly guide me about a method which use potassium dihydrogen phosphate with NaOH to form desired buffer.

I have found a method which use 100 ml 0.1 M KH₂PO₄ and 82.2 ml 0.1 M NaOH to form a buffer of desired pH.

Can anyone guide me how should i make calculation to form 0.067 M potassium phosphate buffer (with NaOH) , pH 7.5.
thanks.

[Borek]

Generally - Henderson-Hasselbalch equation and neutralization stoichiometry is all you need.

If you want result only, having no idea how it was calculated - Buffer Maker.

[kirsch]

Do you have access to a pH meter?  (ie, are you actually doing this for a lab setting, or are you doing it for a classwork assignment?)

If you have potassium dihydrogen phosphate and you add NaOH (in solution), you will convert some of the former to the desired dibasic phosphate form in the solution itself.  Hence, you have your buffer system.  Remember--there is more than one way to prepare a buffer, and using a weak acid/strong base is one way to do it. 

If you have a pH meter, you can (using the NaOH) adjust the solution pH to 7.5.  At that point, you will have the appropriate conjugate base-to-acid ratio of dibasic phosphate to monobasic phosphate for a proper buffer system.  It has to be that way to attain that pH value (see the HH equation). 

However, I can't tell you (from that method) exactly what your concentrations will be.  For that, as Borek said, you'd need to do some calculations with HH equation. 

One last thing--if you must do this on paper, you may need to account for total ionic strength and temperature.     

[mehc]

Thanks for both replies. They really helped me a lot.

I have access to pH meter. I want to determine tea polyphenols by ferrous tartrate method. And in the method I follow, the authors mentioned the use of 0.067 M potassium dihydrogen phospate buffer pH 7.5.

As dibasic phospate will form in solution, how can I use Henderson-Hasselbalch equation to determine concentrations?

If I use the equation with NaOH as base, i think it will not give valid results, as Henderson-Hasselbalch equation is for weak acids and bases, and NaOH is strong base.

If I make such buffer by preparing 0.067 M KH₂PO₄, and adjusting it pH with NaOH to 7.5, will it be correct?

[Borek]

H₂PO₄^- is your acid and HPO₄^2- is your base - you add some hydroxide to get to the correct ratio of both. This ratio is described by HH equation. If you titrate with NaOH till you get the correct pH you are doing exactly the same, just experimentally.

[mehc]

ok, Thanks a lot.

[kirsch]

Key in on the first sentence Borek said in his last post. 

NaOH is a strong base--that's true.  But you're not using IT as the conjugate base in your buffer system.  You're only using NaOH to produce the conjugate base of the monbasic phosphate...ie, (K+)H2PO4 --> (K+)(Na+)(HPO4) 

I mean, you could go buy monobasic and dibasic phosphate, use the HH equation, and figure out how much of each to add to solution.  But you can also prepare what you need in solution with the help of a strong base (in this particular case). 

As for calculations with HH, remember, the "log part" is a RATIO.  Thus, let's say you start with this value of 0.067 M monobasic phosphate (potassium dihydrogen phosphate).  That's your acid.  So, the equation is "plug and chug" from there.  You know your desired pH is 7.5.  You can look up the pKa.  (Remember to choose the right one since phosphate is polyprotic).  All you need to solve for is conjugate base, which is the concentration of dibasic phosphate.  HOWEVER--if you don't adjust for ionic strength and temperature, your calculations will be off a little.   

But, the reason I point out that it is a ratio is that you could start with a different concentration (other than 0.067 M) for monobasic.  And, then, you would get a different value for dibasic to keep the pH at 7.5.

Anyhoo--you can bypass all the calculations with a pH meter, because it does the work for you "behind the scenes."  You can be "experimental" about it.  Kind of like "guess and check" vs "algebra."   So, in response to your last sentence, mehc, yes, adjusting it to pH 7.5 will be correct, because it has to be.  That's the rules of chemistry.  The HH equation is based around those rules.  The pH won't be 7.5 unless there is the correct ratio of mono and dibasic phosphate.  And that ratio has to account for various things, including ionic strength and temperature.

[mehc]

Thanks. I am really greatful to both of you.