Question

Enter your answer in the provided box. How many milliliters of 1.98 M NaOH must be added to 2.00 x 102mL of 0.20 M NaH,PO, to make a buffer solution with a pH of 6.90? mL 12 and available

Answer 1

Ka for H2PO4 = 6.2 x 10-8 pKa=-log(6.2 x 10-8) = 7.21 Let V (in L) be the volume of NaOH to be added. Number of moles of NaOH = M*V = 1.98 M*VL = 1.98V mol Total volume = 207 mL = 0.207 L Volume of NaH PO, acid = 2.00 x 102 mL = 0.2 L Number of moles of NaH,POA = M*V = 0.20 M* 0.2 L = 0.04 mol Net moles of acid remaining = 0.04 – 1.98V mol NaH2PO4 + NaOH = Na HPO4 + H20 Initial 0.04 0 Add -- 1.98V Change -1.98V -1.98V +1.98V Equilibrium 0.04 -1.98V 1.98V [H2PO4 B - net moles Total volume net moles (0.04–1.98V) mol 0.2+V 1.98V mol [HPO4 FTotal volume 0.2+V According to Henderson equation, [base] pH = pka+log 10% [acid] 1.98V mol 6.90 = 7.21+log – 0.2+V (0.04-1.98V mol 0.2+V 1.98V 6.90 = 7.21+log – * (0.04 – 1.98V) log_ 1.98V = 106.31 -=-0.31 (0.04 -1.98V) 1.98V (0.04 -1.98V 1.98V -=0.49248 (0.04 -1.98V) 1.98V = 0.019699-0.9751178V 2.955117V = 0.019699 V = 0.006666L = 6.67 mL volume of NaOH to be added = 6.67 mL