How many milliliter a of 1.0 M NaOH must be added to a 200 mL of...

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Howmany milliliter a of 1.0 M NaOH must be added to a 200 mL of 0.10 MNaH2PO4 to make a buffer solution with a pH of 7.50?

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3.8 Ratings (412 Votes)

200 mL* 0.10 M H2PO4^- = 200*0.1 = 20 millimols.
----mL* 1.0 M NaOH = ---- millimoles.

H2PO4^- + OH^- ==> HPO4^- + H2O
20Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â initial

-x Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â change
20-xÂ Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â equilibrium

pH = pKa + log[(base)/(acid)]

7.50 = 6.86 + log(x/20-x)

x = 16.27 mL

7.50 = 6.86 + log(x/20-x)

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