Question

A buffer solution contains 0.308 M NaHCO, and 0.256 M K CO Determine the pH change when 0.066 mol HT is added to 1.00 L of the buffer. pll change

Answer 1

str 3 NaHCO3 is ong electr olyte.so it fully dissociaters as NaHCO3lar) Natag + HCO3(aq) So, [NaHCO3] [HCO3 ] = 0.308m om the othes hand Kz Coa also str ong electrolyte f fully dissociates Kzco Coz has 2Kt (aq) + con lag) CK2CO3 [cos 2-] = 0.256 Here HCO3 act weak acid dissociates CO32- fag) Hom (as) ² H+ (aq) + pka of it com = 10.32 CO2 is conjugate base of Hcos as gr as

weak acid will act can In the solution (HC0g) & it's conjugate bare (eeg?) both are foresenta So, the solution buffer For buffer system we apply Hanel balch equation the Henderson where, I TA Ana Acid here, pka = 10:32 CB = conjugate A] > [H603] = 0.308M [B] E CO32- FC B] pHapkatlog f bare 0.256 M 0.308 pH= 10:32 + log fors6 Tio 0008 10.32 = 1024 (berforre adding to Initial pH 10.rul ," 0.308 M -- 7 Buffer volume is 12f initial CHCO3 & [ CO32-1 0.256 m Im in buff buffer molen & com present 01308 mol & 0:2 stimarer pectivory of Hcom