Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH.

2.1×10−4 M KOH

4.9×10−4 M Ca(OH)2

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• david
Lv 7
2 months ago

2.1×10−4 M KOH

KOH  -->  K+  +  OH-  <<<  notice 1 KH forms 1 OH- ... same numbers

2.1×10−4 M KOH   forms  2.1×10−4 M [OH -]

[H3O+][OH-]  =  1x10^-14

[H3O+]  =  4.76x10^-11 M

pH = -log[H3O+]  = 10.32

pOH = -log[OH-]  =  3.68

====================================4.9×10−4 M Ca(OH)2

Ca(OH)2  -->  Ca2+  +  2OH-   <<<<  OH- is 2 times  Ca)OH)2

[OH-]  =  9.8x10^-4 M

[H30+]  =  1x10^-14 / 9.8x10^-4  =  1.02x10^-11 M

pH = 10.99

pOH = 3.01

• 2 months ago

Both calculations go exactly the same way as in

A)

[OH−] = 2.1×10^−4 M

[H3O+] = 10^-14 / (2.1×10^−4) = 4.8×10^−11

pH = - log (4.8×10^−11) =  10.3

pOH = - log (2.1×10^−4) =  3.7

B)

Ca(OH)2 → Ca{2+} + 2 OH{-}

Supposing complete dissociation:

[OH-] = 2 x (4.9×10^−4 M) = 9.8×10^−2 M

[H3O+] = 10^−14 / (9.8×10^−2) = 1.02×10^−13 M

pH = - log (1.02×10^−13) = 13.0

pOH = - log (9.8×10^−2) = 1.0