Question

A microbiologist is preparing a medium on which to culture E. coli bacteria. She buffers the medium at pH 7.00 to minimize the effect of acid-producing fermentation. What volumes of equimolar aqueous solutions of K₂HPO₄ and KH₂PO₄ must she combine to make 750.0 mL of the pH 7.00 buffer?

K_a values for phosphoric acid:

K_a1 = 7.2 ×10⁻³

K_a2 = 6.3 ×10⁻⁸

K_a3 = 4.2 × 10⁻¹³

Volume H₂PO₄⁻ = ??? mL

Volume HPO₄²⁻ = ??? mL

Answer 1

The Ka values for the phosphoric acid are The phosphoric acid has the following ionizations Ka1 = 7.2 × 10-3 Ka2-6.3 x 108 pKa2log(6.3 x 10-8) 7.20 pka, = 4.2 × 10-13 H2PO4HPO42H Since the given pH (7.00 ) is nearer to second ionization. The phosphate buffer must be prepared by H2PO and HPO42 pH- pk,+log [base] [acid] HPO,2] 7.00-1.20-10g H,Po, HPo,2 log -0.20 100200.63 HPO,] HPO,1-0.63x[H,PO,] Let the conc. of buffer 1.0M concentrations of the solutions 1.0 M That is [HPO4-2 ] + [H.PO,-] = 1.0 1.63x[H,PO,1-1 .630.613497 [HPO,1-0.63x[H,PO,10.63x0.613497 0.386503 [H,PO,]-0.613497AM HPO0.386503M Volume of the buffer solution 750 mL = 0.75 L Number of moles of H,PO,-present in the solution-M *V-0.6 13497M×0.75 L-0.460 123 mol Number of moles of HPO present in the solution M*V 0.386503Mx0.75 L 0.289877 mol 1.00 M-_-: 460.1 mL 0.289877 mol = 0.289877 L =289.9 mL Vohume of KH,PO, soutionmoles0.460123 mol moles Volume of K,HPO, solution 1.00 M molarity 1