Question

375 mL of a 0.88 M potassium hydroxide solution is added to 496 mL of a 0.76 M cesium hydroxide solution. Calculate the pOH of the resulting solution.

Answer 1

pOH of resulting solution is 0.086

Step-by-step explanation:

KOH and CsOH are monoacidic strong base

Number of moles of
`OH^(-)` in 375 mL of 0.88 M of KOH =
`(0.88* 375)/(1000)moles` = 0.33 moles

Number of moles of
`OH^(-)` in 496 mL of 0.76 M of CsOH =
`(0.76* 496)/(1000)moles` = 0.38 moles

Total volume of mixture = (375 + 496) mL = 871 mL

Total number of moles of
`OH^(-)` in mixture = (0.33 + 0.38) moles = 0.71 moles

So, concentration of
`OH^(-)` in mixture,
`[OH^(-)]` =
`(0.71)/(871)* 1000M=0.82M`

Hence,
`pOH=-log[OH^(-)]=-log(0.82)=0.086`