Campuran 40 mL larutan NH4OH 0,15 M (Kb = 10^-5) dengan 20 mL larutan HCl 0,3 M mempunyai pH .... A. 4 D. 9 B. 5 E. 10 C. 7

PH CAMPURAN (langsung substitusi rumus) mol NH4OH = M × V mol NH4OH = 0,15 × 40 mol NH4OH = 6 mmol mol HCl = M × V mol HCl = 0,3 × 20 mol HCl = 6 mmol NH4OH + HCL --> NH4Cl + H2O m : 6 6 - - r : 6 6 6 6 -------------------------------------------------------- s : - - 6 6 [NH4Cl] = mol NH4Cl/V total [NH4Cl] = 6 / 40 + 20 [NH4Cl] = 6/60 [NH4Cl] = 0,1 M [H+] = √Kw/Kb × M [H+] = √10¹⁴/10⁵ × 10^-1 [H+] = √10^-¹⁰ [H+] = 10^-5 M pH = -log [H+] pH = -log 10^-5 pH = 5 pH kedua larutan tersebut adalah 5. (B)