\(n_{HCl}=0,1.0,002=0,0002\left(mol\right)\\ n_{NaOH}=0,003.0,1=0,0003\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ Vì:\dfrac{0,0003}{1}>\dfrac{0,0002}{1}\Rightarrow NaOHdư\\ \Rightarrow pH=14+log\left[OH^-\right]=14+log\left[\dfrac{0,0001}{0,1+0,1}\right]=10,69897\)