Question

A buffer solution contains 0.351 M KHSO3 and 0.331 MK S03. Determine the pH change when 0.100 mol KOH is added to 1.00 L of the buffer. pH after addition - pH before addition - pH change = Submit Answer

Answer 1

Molarity of HSO3- Molarity of SO3 2- We know, Moles = Molarity (M) x Volume (V in L) 0.351 M So Moles HSO3- * 1.000 mol 0.3510 0.331 M 0.3510 mol Volume of buffer V = 1000.0 ml = 1.000 L Moles SO3 2- 0.3310 1.000 = (note: 1ml 0.3310 mol 10-3 L) We know pKa -log (Ka) log [ 6.30E-08 ] 7.201 Given Ka2 6.30E-08 = Initial pH of solution: Henderson equation for buffer solution pH= pKa log [Base/Acid] here the Base is SO3 2- and acid is HSO3- log [SO3 2- / HSO3-] log [0.3310 log 0.9430 1 РКа pH 0.3510 7.201 = 7.201 = 0.025 7.201 = = 7.175

OH- addition to buffer : When OH- is added to the buffer, it reacts with acid HSO3- and produce SO3 2-. So during the addition of OH- to the buffer; SO3 2- Conc/moles will increase and HSO3- moles / conc will decrease as shown in the below table. 0.1 so Moles of OH- added mol = HSO3- (aq) + ОН- (аq) SO3 2- (аq) + H20 () Initial 0.351 0.1000 0.331 - 0.100 Change Equilibrium - 0.100 0.100 0.2510 0 0.4310 og [SO3 2-/ HSO3-] log 0.4310 log 1.7171] РКа pH 0.2510 7.201 = 7.201 0.2348 7.201 = = 7.435 pH change pH after addtion - pH before addition 7.435 7.175 0.260