Answer
$$ \boxed{0.02125 M H^+}$$
Work Step by Step
Since we know the molality of both solutions and the fact that
$$H^{+} + OH^{-} \rightarrow H_2O$$
We can simply calculate which is in excess, as the limiting reagent will determine how much water is made.
Thus we find the moles in the solution for each,
For $HCL_{aq}$
$$\frac{0.25}{1} mol/L * \frac{75}{1000} L = 0.01875 mol/L $$
And for barium hydroxide,
$$\frac{0.055}{1} mol/L * \frac{225}{1000} L = 0.012375 mol/L $$
Subtracting gives,
0.006375 moles. of excess $H^+$
and the total volume is 300 mL.
$$0.006375 moles / 0.3 L = 0.02125
$$
or$$ \boxed{0.02125 M H^+}$$
