解：设100.0 g废液中$\mathrm {Ba}(\mathrm {OH)}_{2}$的质量为$x。$$ $
$\mathrm {Ba}(\mathrm {OH)}_2+\mathrm {H}_2\mathrm {SO}_4\xlongequal[ ]{ }\mathrm {BaSO}_4↓+2\mathrm {H}_2\mathrm {O}$
171 233
$x$ 2.33 g
$\frac {171}{233}=\frac {x}{2.33 \mathrm {g}}，$解得$ x=1.71 \mathrm {g} $
废液中$\mathrm {Ba}(\mathrm {OH)}_{2}$的质量分数为$\frac {1.71 \mathrm {g}}{100.0 \mathrm {g}}×100％=1.71 ％$
答：废液中$\mathrm {Ba}(\mathrm {OH)}_{2}$的质量分数为$1.71\%。$$ $
