\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\)
\(n_{CuCl_2}=\dfrac{13,5}{135}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
_____0,3---------------->0,3
2NaOH + CuCl2 --> Cu(OH)2 + 2NaCl
_0,2<----0,1
=> nNaOH dư = 0,1(mol)
=> \(C_{M\left(OH^-\right)}=\dfrac{0,1}{1}=0,1M\)
=> pH = 14 + log(0,1) = 13
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nCuCl2=13,5135=0,1(mol)nCuCl2=13,5135=0,1(mol)
PTHH: 2Na + 2H2O --> 2NaOH + H2
_____0,3---------------->0,3
2NaOH + CuCl2 --> Cu(OH)2 + 2NaCl
_0,2<----0,1.
=> nNaOH dư = 0,1(mol)
=>
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