Question

Use the Henderson-Hasselbalch equation to calculate the pHpH of each solution:

A)

a solution that is 0.175 MM in HC2H3O2HC2H3O2 and 0.125 MM in KC2H3O2

B)

a solution that is 0.190 MM in CH3NH2CH3NH2 and 0.135 MM in CH3NH3Br

Answer 1

A) [M (2H3 02] = 0.175m [15 C2H302] = 0.125m (ch3 cool + CH3Coolc) is an acidie buffer whose pH is given by Henderson - Hasselbalch equations PH = PEGG + log | Metsor where pke = -lopka & ka af HC2M302= 1-8810 = pka = -2; 1.8x155 = 4.745 - pH = 4.745 tlog 0.125 = 4.745-0.146 - PH = (4.598 B) [CM3 NH?] =0-190m (CH 3 W M3 Br] = 0-135m {eng Nhz + CH3 NM3 Borl is a basee buffer whose PH 5 given as [engning for where pkg=_bog ky - and Ich for CH3 WW2 = 44 810 :. Pk6= -2; 4.4x10" = 3.36 Pon= 3.36 +log 0-135 Pok= 3-36-0-198=3.212 PH=lu-Pon RH=14-3.212 = 10.788 - PH = 10.788 pon = pky tlog s cho nh27 0.190