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)
Hey there!:
Molar mass H3PO4 = <span>97.9952 g/mol
Atomic Masses :
H = </span><span>1.00794 a.m.u
</span>P = <span>30.973762 a.m.u
</span>O = 15.9994 a.m.u<span>
H % = [ ( 1.00794 * 3 ) / </span> 97.9952 ] * 100
H% = <span>3.0857 %
P % = [ ( </span>30.973762 * 1 ) / 97.9952 ] * 100
P% = <span>31.6074 %
O % = [ ( </span>15.9994 * 4 ) / 97.9952 ] * 100
O% = <span>65.3069 %
Hope this helps!</span>
Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature