how many millilitresof na2hpo4(c =1 mol/l) and nah2po4(c= 1mol/l) should be taken to prepare 450 ml of buffer with ph =6.4 pk(h2po4)=6.8

There is a formula to calculate pH of buffer made of weak acid (here it's NaH2PO4) and its salt with strong base (Na2HPO4):

pH = pKa+lg CM Na2HPO4/CM NaH2PO4;

let's take that 450 ml of buffer is sum of volumes of Na2HPO4 and NaH2PO4 solutions (V1 ml - volume of Na2HPO4 sol. and V2 ml - volume of NaH2PO4 sol.), i.e. V1+V2 = 450 ml = 0.45 l. (1);

as CM = n/V and CM both of salts = 1 mol/l, so we have following: CM Na2HPO4 = V1/0.45 = 2.222*V1 and CM NaH2PO4 = V2/0.45 = 2.222*V2;

lg 2.222*V1/2.222*V2 = pH-pKa = 6.4-6.8 = -0.4, so 2.222*V1/2.222*V2 = 10^-0.8 = 0.3981;

as V2 = 0.45-V1 (see (1) above), we get 2.222*V1/2.222*(0.45-V1) = 0.3981, so V1 = 0.128 l. or 128 ml;

V2 = 0.45-0.128 = 0.322 l. or 322 ml;

Volume of Na2HPO4 sol. is 128 ml.;

Volume of NaH2PO4 sol. is 322 ml.

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