$-nNaOH=0,012.0,1=0,0012 (mol)$
$NaOH+HCl→NaCl+H_2O$
0,0012→0,0012 0,0012 (mol)
$-Vdd_{HCl}=\frac{0,0012}{0,01}=0,12 (l)$
a.
-Dd X gồm:NaCl
-Vdd sau pứ=0,1+0,12=0,22 (l)
⇒$[NaCl]=\frac{0,0012}{0,22}=\frac{3}{550}(M)$
$NaCl→Na^++Cl_-$
$[Na^+]=[Cl^-]=\frac{3}{550} (M)$
b.
$Ta có: [NaOH]=0,012 (M)$
$⇒[OH^-]=0,012 (M)$
$⇒pH=14+log[OH^-]=14+log(0,012)≈12,079 $
--------------------Nguyễn Hoạt-----------------
