Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of
in 375 mL of 0.88 M of KOH =
= 0.33 moles
Number of moles of
in 496 mL of 0.76 M of CsOH =
= 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of
in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of
in mixture,
= 
Hence, ![pOH=-log[OH^{-}]=-log(0.82)=0.086](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D%3D-log%280.82%29%3D0.086)
Answer: 3.75 M
Explanation:
400 mL = 0.4 L
NaOH has a molar mass of around 40 g/mol.
= 1.5 moles
Molarity =
= 3.75 M