Pembahasan soal kimia jika 40 ml NaOH 0,1M dicampur dengan 10ml H2SO4 0.1M tentukan pH campuran


2NaOH + H2082SO2084 —> Na2082SO2084 + 2H2082O, m 4mmol 1mmol – -, b 2mmol 1mmol 1mmol 2mmol, s 2mmol – 1mmol 2mmol, mol sisa NaOH = 2mmol, M = 2/50 = 4x10207b\xb2, NaOH —-> Na207a+OH207b, 1 1 1, OH207b = 1 x 4 x10207b\xb2 = 4x10207b\xb2, pOH = -log4x10207b\xb2 = 2-log4, pH = 14 – (2 – log4), = 12 + log 4, = 12 + 0,6 = ,
,
NaOH –> mol OH-, =M.V.Valensi, =0,1.40.1, =4, H2SO4 –> mol H+, =M.V.Valensi, =0,1.40.2, =8, Sisa mol H+ = 8-4 =4, H+sisa = mol sisa H+ per V.Campuran, =4/40 = 0,1, pH = -log 1 x10 pangkat -1 =, = 1