Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of [tex]OH^{-}[/tex] in 375 mL of 0.88 M of KOH = [tex]\frac{0.88\times 375}{1000}moles[/tex] = 0.33 moles
Number of moles of [tex]OH^{-}[/tex] in 496 mL of 0.76 M of CsOH = [tex]\frac{0.76\times 496}{1000}moles[/tex] = 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of [tex]OH^{-}[/tex] in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of [tex]OH^{-}[/tex] in mixture, [tex][OH^{-}][/tex] = [tex]\frac{0.71}{871}\times 1000M=0.82M[/tex]
Hence, [tex]pOH=-log[OH^{-}]=-log(0.82)=0.086[/tex]
