[tex]1.43 \; \text{ml}[/tex]
Explanation
- Number of moles of chloride ions in the final solution: [tex]n = c \cdot V = 0.012 \times 0.300 = 0.0036 \; \text{mol}[/tex]
- Each mole of barium chloride contains two moles of chloride ions. Thus the concentration of chloride ions in the initial 1.51 M barium chloride solution: [tex]\begin{array}{lll} c(\text{Cl}^{-}) &=& 2 \times c(\text{BaCl}_2)\\& =& 2.52 \; \text{mol} \cdot \text{L}^{-1} \\ & = &0.00252 \; \text{mol}\cdot \text{mL}^{-1} \end{array}[/tex]
Therefore
[tex]\begin{array}{lll} V & = & n /c \\ & = & 0.0036 / 0.00252\\ & = & 1.43 \; \text{mL}\end{array}[/tex]
