\(n_{Ba\left(OH\right)_2}=0,075.0,1=0,0075\left(mol\right)\\ 2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+2H_2O\\ n_{HCl}=0,0075.2=0,015\left(mol\right)\\ C_{MddHCl}=\dfrac{0,015}{0,05}=0,3\left(M\right)\)
2HCl+Ba(OH)2->BaCl2+2H2O
n Ba(OH)2=0,1.0,075=0,0075 mol
=>n HCl=0,015 mol
=>CmHCl=0,015\0,05=0,3M