HCl + NaOH → NaCl + H2O
\(n_{HCl}=0,02\times0,1=0,002\left(mol\right)\)
Theo pT: \(n_{NaOH}=n_{HCl}=0,002\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\frac{0,002}{0,01}=0,2\left(M\right)\)
Vậy \(x=0,2\left(M\right)\)
nH+ = 0.002 mol
nOH- = 0.01x mol
H+ + OH- => H2O
0.002_0.002
<=> 0.01x = 0.002
<=> x = 0.2