Đáp án:

\(C{\% _{KOH}} = \dfrac{{74,2}}{{400 + 350}} \times 100\%  = 9,89\% \)

\(C{M_{KOH}} = \dfrac{{1,325}}{{0,68182}} = 1,943M\)

Giải thích các bước giải:

\(\begin{array}{l}
{n_{KOH}}(11,2\% ) = \dfrac{{400 \times 11,2}}{{100 \times 56}} = 0,8mol\\
{n_{KOH}}(8,4\% ) = \dfrac{{350 \times 8,4}}{{100 \times 56}} = 0,525mol\\
 \to {n_{KOH}} = 0,8 + 0,525 = 1,325mol\\
 \to {m_{KOH}} = 74,2g\\
 \to C{\% _{KOH}} = \dfrac{{74,2}}{{400 + 350}} \times 100\%  = 9,89\% \\
{V_{KOH}} = \dfrac{{400 + 350}}{{1,1}} = 681,82 = 0,68182l\\
 \to C{M_{KOH}} = \dfrac{{1,325}}{{0,68182}} = 1,943M
\end{array}\)