To calculate the volume of 0.125 M HCl needed to neutralize 51.0 mL of 0.100 M Ba(OH)₂, we first need to write the balanced chemical equation for the neutralization reaction:
\[ \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]
From the balanced equation, we see that one mole of Ba(OH)₂ reacts with two moles of HCl.
Next, we can determine the number of moles of Ba(OH)₂ in 51.0 mL of 0.100 M Ba(OH)₂:
\[ \text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume(L)} = 0.100 , \text{mol/L} \times 0.0510 , \text{L} = 0.00510 , \text{mol} \]
According to the stoichiometry of the reaction, since 1 mole of Ba(OH)₂ reacts with 2 moles of HCl, the number of moles of HCl needed is:
\[ \text{Moles of HCl} = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 0.00510 , \text{mol} = 0.01020 , \text{mol} \]
Now, we need to find the volume of 0.125 M HCl required to provide 0.01020 moles:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.01020 , \text{mol}}{0.125 , \text{mol/L}} = 0.0816 , \text{L} \]
To convert this volume from liters to milliliters, we multiply by 1000:
\[ \text{Volume (mL)} = 0.0816 , \text{L} \times 1000 = 81.6 , \text{mL} \]
Rounding to three significant digits, the volume of 0.125 M HCl needed is:
\[ \boxed{81.6} , \text{mL} \]