\(V_{ddA}=0,1+0,1=0,2\left(l\right)\\ n_{OH^-}=0,5.0,1.2+0,5.0,1=0,15\left(mol\right)\\ n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,5=0,05\left(mol\right)\\ n_{K^+}=n_{KOH}=0,1.0,5=0,05\left(mol\right)\\ \left[K^+\right]=\dfrac{0,05}{0,2}=0,25\left(M\right);\left[Ba^{2+}\right]=\dfrac{0,05}{0,2}=0,25\left(M\right);\left[OH^-\right]=\dfrac{0,15}{0,2}=0,75\left(M\right)\)