Question

Prepare a buffer by acid-base reactions. Consider how to prepare a buffer solution with pH = 7.35 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.362-M solution of weak acid with 0.325 M potassium hydroxide. pk Weak Acid HNO2 HCIO HCN Conjugate Base NO2 CIO CN K 4.5 x 10-4 3.5 x 10-8 4.0 x 10-10 3.35 7.46 9.40 How many L of the potassium hydroxide solution would have to be added to the acid solution of your choice?

Answer 1

given pH = 7.35 ; so the pKa must be very near to pH

pKa= 7.46 is very near , so buffer could be HClO and ClO-

We know, Moles = Molarity (M) x Volume (V in L) So Moles HCIO= 0.362 * 1.000 mol = 0.362 mol Let moles of OH- to be added = x mol Molarity of Cio- = 0.000 Moles of HCIO = 0.362 Volume of buffer V = 1000 = 1.000 note: 1ml = 10^-3 L M M L mL ; pH of buffer solution = 7.35 When OH- is added, it reacts with acid HClO and produce CIO-. So during the addition of OH-; CIO- will increase and HClO moles / conc will decrease as shown in the below table. HCIO (aq) + OH- (aq) <-----> CIO- (aq) + H20 (1) Initial 0.362 0.000 Change - X +x Equilibrium (0.362-x) 0 Henderson equation for buffer solution is pH=pKa + log [Base/Acid] here the Base is CIO- and acid is HCIO 7.35 = 7.460 + log (x 10.362-x) 7.35 - 7.460 = log (x 10.362-x) -0.110 = log (x /0.362-x) log (x /0.362-x) = -0.11000 (x /0.362-x) = 104 -0.11000) (x/0.362-x) = 0.7762 x = 0.7762 (0.362-x) x = 0.2810 - 0.7762 x x + x 0.7762 = 0.2810 1.7762 x = 0.2810 x = 0.158 Here x = moles of KOH added = 0.158 mol Volume of NaOH = Mole of NaOH/Molartiy of NaOH: 0.158 mol = 0.325 mol L-1 = 0.487