Gọi thể tích dung dịch là a (lít)
\(\left\{{}\begin{matrix}n_{NaOH}=1,2a\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,8a\left(mol\right)\end{matrix}\right.\)
\(n_{HNO_3}=\dfrac{200.12,6\%}{63}=0,4\left(mol\right)\)
PTHH: \(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
1,2a----->1,2a
\(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
0,8a------->1,6a
=> 1,2a + 1,6a = 0,4
=> a = \(\dfrac{1}{7}\left(l\right)\)