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Sagot :
An aqueous solution of a strong base has a pH of 10. 220 at 25°C then concentration of the base if the base is
(a) LiOH = 1.66×[tex]10^{-4}[/tex] M
(b) [tex]Ba(OH)_{2[/tex] = 3.32×[tex]10^{-4}[/tex]M
Calculation ,
For calculation of concentration of the LiOH and [tex]Ba(OH)_{2[/tex] first we have to calculate pOH of the solution.
pH + pOH = 14
pOH = 14 - pOH = 14 - 10.22 = 3.78
pOH = -㏒ [ [tex]OH^{-}[/tex] ]
[ [tex]OH^{-}[/tex] ] = [tex]10^{-pH}[/tex] = [tex]10^{-3.78}[/tex] = 1.66×[tex]10^{-4}[/tex]
So , concentration of the base LiOH = 1.66×[tex]10^{-4}[/tex]
So , concentration of the base [tex]Ba(OH)_{2[/tex] = 1.66×[tex]10^{-4}[/tex] ×2 = 3.32×[tex]10^{-4}[/tex]
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