Question

calculate the concentration of each of the ions in a mixed solution in which no reaction occurs when 125 mL of 0.0240 M strontium sulfide combines with 75.0 mL of 0.0480 M strontium chloride.

Answer 1

1. Find moles of SrS

$$0.125 L\times\frac{0.0240 mol}{L}=0.003 mol$$

$$SrS(aq) \rightarrow Sr^{2+}(aq)+S^{2-}(aq)$$

$$n(SrS)= n(Sr^{2+})=n(S^{2-})=0.003 mol$$

2.Find moles of SrCl2

$$0.075 L \times \frac{0.0480 mol}{L} = 0.0036 mol$$

$$SrCl_2(aq)\rightarrow Sr^{2+}(aq) + 2Cl^-(aq)$$

$$n(SrCl_2)=n(Sr^{2+})=0.0036 mol$$

$$n(Cl^-)=0.0036 mol(SrCl_2)\times \frac{2 mol(Cl^-)}{1mole (SrCl_2)}=0.0072 mol(Cl^-)$$

3.Find total volume of a new solution:

$$V_{total} = 0.125L +0.075 L = 0.2 L$$

4.Find concentration of each of the ions

$$c(S^{2-})=\frac{0.003 mol}{0.2 L}= 0.015 \frac{mol}{L}$$

$$c(Cl^-)=\frac{0.0072 mol}{0.2 L}= 0.036 \frac{mol}{L}$$

$$c(Sr^{2+})=\frac{0.003mol+0.0036mol}{0.2 L}= 0.033 \frac{mol}{L}$$