\(n_{H^+\left(bđ\right)}=n_{HCl}=0,3\left(mol\right);n_{OH^-}=2.n_{Ca\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\Rightarrow H^+dư\\ n_{H^+\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ \left[H^+_{dư}\right]=\dfrac{0,1}{1}=0,1\left(M\right)\\ \Rightarrow pH=-log\left[H^+_{dư}\right]=-log\left[0,1\right]=1\)