Question

A buffer solution contains 0.384 M NaHCO3 and 0.367 M Na,CO3. Determine the pH change when 0.093 mol KOH is added to 1.00 L of the buffer. pH after addition - pH before addition = pH change=

Answer 1

We know, Moles = Molarity (M) x Volume (V in L) Molarity of HCO3- = 0.384 M So Moles HCO3- = 0.3840 * 1.000 mol Molarity of CO32- = 0.367 M = 0.3840 mol Volume of buffer V = 1000.0 ml Moles CO3 2- = 0.3670 *' 1.000 = 1.000 L = 0.3670 mol (note: 1ml = 10^-3 L) Given Ka = 5.60E-11 We know pka = -log (Ka) = -log 5.60E-11 ] = 10.252 Initial pH of solution: Henderson equation for buffer solution pH=pKa + log [Base/Acid] here the Base is CO32- and acid is HCO3- pH = pka + log (CO3 2- / HCO3- ] = 10.252 + log [ 0.3670 0.3840 ] = 10.252 + log [ 0.9557 ] = 10.252 + -0.02 10.23

OH- addition to buffer : When OH- is added to the buffer, it reacts with acid HCO3- and produce CO3 2-. So during the addition of OH- to the buffer; CO32- Conc/moles will increase and HCO3- moles / conc will decrease as shown in the below table. so Moles of OH- added = 0.093 mol HCO3- (aq) + OH- (aq) <-----> CO3 2- (aq) + H20 (1) Initial 0.384 0.0930 0.367 Change - 0.093 - 0.093 + 0.093 Equilibrium 0.2910 0.4600 pH = pka + log [CO3 2- / HCO3-] = 10.252 + log [ 0.4600 = 0.2910 ] = 10.252 + log [ 1.5808 ] = 10.252 + 0.1989 10.451 pH change = pH after addtion - pH before addition = 10.451 - 10.232 = 0.22