How would I make 100mL of a .100 M phosphate buffer with a pH of 7.2 using 1 M NaH2PO4 and Na2HPO4 solid (MW= 142 g/mol)?

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  1. pH = pK2 + log (base/acid)
    I looked up pK2 and found 7.2; therefore,
    7.2 = 7.2 + log (base/acid)
    so (base/acid) = 1
    If we used 100 mL of the 1 M NaH2PO4, that is 1 M for the concn of acid.
    If we used 0.1 mole Na2HPO4 in 100 mL that is 0.1 mole/0.1 L = 1 M so the ratio is 1.
    0.1 mole Na2HPO4 must be 14.2 grams.

    Will that do it? We can see.
    pH = 7.2 + log (base/acid)
    pH = 7.2 + log (1 M base/1 M acid)
    pH = 7.2 + 0 = 7.2
    Technically this will make slightly more than 100 mL because 100 mL solution + 14.2 g of the salt will likely increase the volume slightly.

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