Question

How many milliliters of a 70.0 ml solution of 1.51 m bacl2 must be used to make 12.0 ml of a solution that has a concentration of 0.300 m cl–?

Answer 1

$1.43 \; \text{ml}$

Explanation

• Number of moles of chloride ions in the final solution: $n = c \cdot V = 0.012 \times 0.300 = 0.0036 \; \text{mol}$

• 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: $\begin{array}{lll} c(\text{Cl}^{-}) &=& 2 \times c(\text{BaCl}_2)\\& =& 2.52 \; \text{mol} \cdot \text{L}^{-1} \\ & = &0.00252 \; \text{mol}\cdot \text{mL}^{-1} \end{array}$

Therefore

$\begin{array}{lll} V & = & n /c \\ & = & 0.0036 / 0.00252\\ & = & 1.43 \; \text{mL}\end{array}$