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Sagot :
Answer:
pH= 14.17
pOH= -0.17
[H3O+]= 6.76x10^-15M
[OH-]= 1.5M
Explanation:
1st) It is necessary to write the dissociation equation:
[tex]Al(OH)_3\rightarrow Al^{+3}+3OH^-[/tex]From the equation we can see that from 1 mole of Al(OH)3, 3 moles of OH- are dissociated. So, in this case, from the 0.5M solution, 1.5M of OH- will be produced (3x0.5M=1.5M).
We know that the concentration of OH- is [OH-]= 1.5M.
2nd) With the [OH-] we can calculate the pOH of the solution:
[tex]\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ pOH=-log(1.5M) \\ pOH=-0.17 \end{gathered}[/tex]The pOH is -0.17.
3rd) Now we can calculate the pH of the solution:
[tex]\begin{gathered} pH+pOH=14 \\ pH=14-pOH \\ pH=14-(-0.17) \\ pH=14.17 \end{gathered}[/tex]The pH is 14.17.
4th) Finally, we can calculate the concentration of H+:
[tex]\begin{gathered} 14.17=-log\lbrack H^+\rbrack \\ 10^{(-14.17)}=\lbrack H^+\rbrack \\ 6.76*10^{-15}M=\lbrack H^+\rbrack \end{gathered}[/tex]So, the concentration of H+ is 6.76x10^-15M.
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