Question

5. What is the pH of 50 mL of a 1.0 mol/L methylamine (CH3NH2) solution after the addition of 50 mL of a 0.5 mol/L solution of HCI? Your angwer

Answer 1

+ (6) When Hel is added to methylamine solution then CH₃NH3 el salt is produced The reaction is shown below CH₃NH₂ Hal > CH₃NH3 d concentration of CH3NH2 50 ml x 1.0 m (50 ml + 50 ml) 0.5 m concentration of Hel 50 ml X 0.5 m (50+50 ml 0.25 m .. 0.25 m Hell will react with 0.25 m CH3NH2 to produce 0.25 m CH3NH3 el salt. After reaction , Satt produce, [CH3NH3 el] = 0.25 m [C H3NH2] - (0.5-0.25) m = 0.25 m so it will acts as basic buffer. According to Henderson's equation. pOH = pkle + log [CH3NH3 el] I CH3NH₂ pk le for methylamine is 3.31 at 25°c. рон 3. 34+ log Oms 0.25 poh = 3.34+ log! pot [ log 1=0 We know that a pH + POH = 17 Base remain, -334 .PH IT-POH PH 14-3:34 10.66 of this solution will be 10.66. PH= PH