Question

A buffer that contains 0.27 M of a base, B and 0.44 M of its conjugate acid BH⁺, has a pH of 9.05. What is the pH after 0.03 mol of NaOH are added to 0.79 L of the solution?

Answer 1

Anstoeri- Given, Buffeo Solution contains. Base - 0 WFM. conjugate acid = 0.44M :PH: 9.05 PH aftem:0.03 mol Of NaOH. Solution=oyal. Base y amount of solteon 502780.79 10.2133 mola congugate acid x amount of Solection -0.4480.79=0.34 760 mol, After 0.03 moi, The change in equilibrium is 0.9433 and 0.31760 These values are obtained from list 0.2133+ 0.03 0-2433 0.34760 -0.03=0.31760 NOWO, From Hendesen couuation

pot = pkb + log [conjugare acid)/(base] Wiikit pOH = 14-PH = 149.05 - 4.95 Substitute the values in eo 4.958 PKB4109 [0.4410:27 4.95= pkb + 10g [16296296] 4.95 - Pisb+0.28100889 Pkb: 4.95-0.21 PKb = 4.74 Now, we have to find the ph efter adding NaOH. POHE PKB+ log (conjugate acids/base Substiture values after Naoh in ewo POHE 4.747 109 0:3176 10.2433 pot = 4.447 109 11.3053) POH 3 4.74+ 0.115 poh = 4.855.

hlikit PH = 14-POH =14-4.855 29.14 n. The pH after 0.03 mol of Naoh arel I added to 0.79L of solution is 9.14.