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

Answer:

pOH of resulting solution is 0.086

Explanation:

KOH and CsOH are monoacidic strong base

Number of moles of $OH^{-}$ in 375 mL of 0.88 M of KOH = $\frac{0.88\times 375}{1000}moles$ = 0.33 moles

Number of moles of $OH^{-}$ in 496 mL of 0.76 M of CsOH = $\frac{0.76\times 496}{1000}moles$ = 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, $[OH^{-}]$ = $\frac{0.71}{871}\times 1000M=0.82M$

Hence, $pOH=-log[OH^{-}]=-log(0.82)=0.086$