Question

498 mL of a 0.21 M sodium hydroxide solution is added to 158 mL of a 0.61 M rubidium hydroxide solution. Calculate the pOH of the resulting solution.

Answer 1

498 mL of 0.21M Naoh solution molarity (M)= 0.2189 Volume (u) = 498 mL = 498x10*3 L So, moles of Naon = molarity (M) x Volome = 0.21% 498*10-3 pel = 104.58 mmol Naoh - Na + oHo (aq) (aa 104.58 minul 104.58 mmol 104.58 mmol so, moles of OH = 104.58 mmol from Naoh Similarly, moles of RbOH = 0.61% 158 x 10-3 mol = 96.38 mmol RbOH — Rb aq) + oHagy 96.38 mino) 96.38 mmol 96.38 mmol . so, moles of OHO - 96038 mmol from Rbor toral mocy of OH" = (96.38 +104-58) monol =200696 mmol

total volume of Solution aftor = (498+ 158 m2 mixing = 656 me so, concentration of Cora) - mous Volume = 200.96 mmol - 656 m2 m2 a 0. 306 M por= -log [ba -- 1090.306 por = 0.514]