What is the pOH of a 5.6 x 10-5 M solution of cesium hydroxide (CsOH)?

Chemistry

1 answer:

Rudiy273 years ago

7 0

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

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An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

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$V=0.0310L=3.10mL$

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Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

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$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$