Question

HELP!!!

The Henderson-Hasselbalch (Buffer) Equation • The pH of the buffer solution is dependent more on pk, of the buffer than concentrations of acids and bases • As a rule, this equation is only useful if HA and A differ by less than a factor of 10 The K, of HCN is 4.9 x 10 What is the pH of a buffer solution that is 0.100 Min HCN and (0.200 M in KCN? Calculate the pH of a buffer which contains 0.500 Mcaswellic acid and 0.450 M sodium caswellate (The K, of caswellic acid is 4.1 x 10 )

Answer 1

for audic The Henderson - Hasselboleti equation buffer solution can be given as ..pH = pkat log [Salt I [Arid (8) @ BY REN , Ka= 4.9 X10 1 -10 .: pka = -log ka = -log (4.9x10° ) = -log(4.9) – logio -10 = -0.69 +10 = 9.31 Given; [HCN] = 0.1M = [Acid] and [KCN ] = 0.2M = [salt] Thus, pH = pka (HCN) + 9 [HUN] = 9.31+ logorim) = 9.31 + loga = 1PH = 9.316 0:301 9.611) Au

1 for caswellic auid, ka = 41 x1018 pka = -log(4.1 x108) = -log(4-1) - logio 18 =-0.613 + 8 = 7.387 . pkan 7.4 and given; .. and [ caswellic aud] = 0.5M [sodium caswellale] = 0.45m - pH = pka t log [sodium casavellate] Craswellic aud] = = 7.4 + log (9.5 ) boa 10:45M) = 7.4 + log (0.9) . = 7.4 -0.046 = 7.354 ! : Thus, LipH = 7.35