Question

NEL C UCUCU 01 LISUUS How many moles of sodium hydroxide would have to be added to 250 mL of a 0.350 M nitrous acid solution, in order to prepare a buffer with a pH of 3.550? moles

Answer 1

To solve this, we need ka of Nitrous acid. Ka = 4.5 x 10-4 Let n be the number of moles of NaOH to be added. Number of moles of NaOH = n Let volume of NaOH = V (L) Number of moles of HNO, = M*V = 0.350 M* 250 mL = 87.5 mmol = 0.0875 mol Net moles of acid remaining = 0.0875 - n mol HNOZ + NaOH NaNO + H20 Initial Add 0.0875 - n Change -n +n Equilibrium 0.0875 - 0 HNO.)=_net moles_(0.0875-n) mol Total volume (0.250+V)L net moles [NO, ]=- Total volume (0.250+V)L According to Henderson equation, pH = pKa+log [base] PAPA Flos [acid] n 3.550 =- log (4.5x104)+log7 (0.250+V)L (0.0875-n) mol (0.250+V)L 3.550 = 3.347+109 (0.0875 – n) logo" log (0.0875 – n)" =0.203 D -n =100203 (0.0875 n 1.59666 (0.0875 – n) . n=0.139708–1.59666n 2.59666 n=0.139708 V = 0.053803 mol number of moles of NaOH to be added = 0.0538 mol