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I want to prepare an aqueous solution of a mobile phase of $\pu{0.04M}$ Brij-35. The mobile phase has to be adjusted to pH $7.4$ with $\pu{0.5M}$ phosphate buffer which will be prepared with $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ solutions.

Can anyone help me to prepare a phosphate buffer solution of $\pu{0.05M}$ using $\ce{NaH2PO4.2H2O}$ and $\ce{Na2HPO4.7H2O}$ with pH $7.4$?

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closed as off-topic by Mithoron, andselisk, Todd Minehardt, bon, airhuff Oct 16 '17 at 17:52

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    $\begingroup$ @fl.pf. The shorter $\ce{NaH2PO4.2H2O}$ works perfectly fine, mhchem handles quite a lot of shorthand. (I am not saying that that would have been my reason to further edit, but it might help you save some time for future edits. I simply wanted to fix the title in this case.) $\endgroup$ – Martin - マーチン Oct 16 '17 at 11:31
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One possibility is to use the Henderson-Hasselbalch equation $(1)$.

$$\mathrm{pH} = \mathrm pK_\mathrm a + \lg\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}\tag{1}$$

Plugging the $\mathrm pK_\mathrm{a2}$ value of phosphoric acid ($7.198$) and the desired $\mathrm{pH}$ value will give you a ratio of hydrogenphosphate and dihydrogenphosphate concentrations which you can solve to the desired total concentration $c_\text{tot}$ by simple algebra.

However, SigmaAldrich also maintains a nice buffer reference centre that includes ready-calculated masses for different $\mathrm{pH}$ values. All that remains is to make sure your final concentration is the desired one.

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  • $\begingroup$ as you can see using SigmaAldrich they provided calculations using concentration of 0.2M but mine is o.o5M according to my question above, how can I convert concentration from o.2M to o.o5M ? will I still use the same amount of ml provided ? $\endgroup$ – UHAGAZE D. Serge Oct 18 '17 at 1:45
  • $\begingroup$ @UHAGAZED.Serge By dividing by four, obviously. $\endgroup$ – Jan Oct 18 '17 at 6:45
  • $\begingroup$ can you divide concentration (in terms of g/l)by four or you divide volume by four ? $\endgroup$ – UHAGAZE D. Serge Nov 27 '17 at 2:50

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