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`a)`

`n_{H^+}=0,1.12,1.10^{-3}=1,21.10^{-3}(mol)`

`H^{+}+OH^{-}->H_2O`

`1,21.10^{-3}->1,21.10^{-3}(mol)`

`->n_{Ca(OH)_2(2)}={1,21.10^{3-}}/2=6,05.10^{-4}(mol)`

`->[Ca(OH)_2]={6,05.10^{-4}}/{5.10^{-3}}=0,121M`

`b)`

`CaO+H_2O->Ca(OH)_2`

`n_{CaO}=n_{Ca(OH)_2(1)}=0,121.500.10^{-3}=6,05.10^{-2}(mol)`

`->a=m_{CaO}=6,05.10^{-2}.56=3,388(g)`

`c)`

Dd nuoc voi trong co `[OH^-]=2[Ca(OH)_2]=0,242(M)`

`->pH=14+lg(0,242)\approx 13,38`

`a)`

`CaO + H_2O -> Ca(OH)_2`

`Ca(OH)_2 + 2HCl -> CaCl_2 + 2H_2O`

`n_{HCl} = (12,1)/(1000) . 0,1 = 1,21.10^{-3} (mol)`

Theo PT: `n_{Ca(OH)_2} = 1/2 n_{HCl} = 6,05.10^{-4} (mol)`

`=> C_{M(Ca(OH)_2)} = (6,05.10^{-4})/(0,005) = 0,121M`

`b)`

`n_{Ca(OH)_2(bđ)} = 0,121.0,5 = 0,0605(mol)`

Theo PT: `n_{CaO} = n_{Ca(OH)_2} = 0,0605.56 = 03,388(g)`

`c)`

`Ca(OH)_2 -> 2OH^{-} + Ca^{2+}`

`0,121 -> 0,242`

`=> pH = 14 + log(0,242) = 10,384`