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I am wondering if it is possible to know exact amount of acid and base that we can put in together to make a buffer.

I want to make a 0.05 M extraction buffer using disodium hydrogen phosphate (DHP) and sodium dihydrogen phosphate (SDP). I want to make 100 mL of buffer so I worked out that I would require 0.78 g SDP and 0.88 g DHP.

The issue is that it sometimes takes a lot of time by mixing one solution into the other and keep it checking with the pH meter, there has to be some mathematical way to get the exact amounts which would save time and my chemicals.

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Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$.

The Henderson-Hasselbalch equation provides a relationship between the ratio of the two concentrations to the pH: $$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$

Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).

$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$ $$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$ $$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$ $$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$ $$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$ $$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$ $$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$

So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.

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    $\begingroup$ The above derivation of course assumes that $[\ce{A}]\ne0$, which would not make sense anyway. $\endgroup$ – Ben Norris Jan 25 '15 at 21:41
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The simplest and still an accurate way is to start with the acid form of the buffer component weighed out to the desired molarity and adjust the pH with 1N NaoH. If your desired pH is greater than the pKa, then you can expect that it would take quite a bit of NaOH. In this case you should start with 0.05M NaH2PO4. But what is your desired pH? When you talk about making a buffer, you should have information on molarity, volume, pH and pKa ready.

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  • $\begingroup$ You can't weigh something to the desired molarity. You can measure a mass, weigh a force, and calculate a concentration (molarity). Also if you start with the desired concentration, and then add acid/base to adjust the pH, then you are diluting the buffer and end up at a much lower concentration than intended. $\endgroup$ – Martin - マーチン May 30 '17 at 10:37

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