Question 15.14: Calculating the Mass of an Ion Present in a Solution How man......
Calculating the Mass of an Ion Present in a Solution
How many grams of Ca ^{2+} ions are present in an aqueous CaCl _2 solution that contains 9.7 × 10^{22} \textrm{ Cl}^- ions?
The equation that governs the ions present in an aqueous CaCl _2 solution is
CaCl _2 (aq) \longrightarrow Ca ^{2+} (aq) + 2 Cl ^- (aq)
The extent of this dissociation is 100% as CaCl _2 is a strong electrolyte (Sec. 15.2).
Step 1 The given quantity is 9.7 \times 10^{22} \textrm{ Cl}^- ions and the desired quantity is grams of Ca ^{2+} ions present.
9.7 \times 10^{22} \textrm{ Cl}^- ions = ? g Ca ^{2+} ions
Step 2 This is a particles-of-A to grams-of-B problem. The pathway, in terms of Figure 15.3, is
Step 3 The dimensional analysis setup for the calculation is
9.7 \times 10^{22} \cancel{\textrm{Cl}^- \textrm{ ions}} \times \frac{1 \cancel{\textrm{mole Cl}^-}}{6.022 \times 10^{23} \cancel{\textrm{Cl}^- \textrm{ ions}}} \times \frac{1 \cancel{\textrm{mole Ca}^{2+}}}{2 \cancel{\textrm{mole Cl}^-}} \times \frac{40.08 \textrm{ g Ca}^{2+}}{1 \cancel{\textrm{mole Ca}^{2+}}}\quad \quad particles A → moles A → moles B → grams B
Note the number 40.08 in the last conversion factor. The molar mass of Ca ^{2+} ions is the same as the molar mass of Ca atoms, as was explained in the discussion preceding this worked-out example.
Step 4 The result, obtained by combining the numerical factors is
\frac{9.7 \times 10^{22} \times 1 \times 1 \times 40.08}{6.022 \times 10^{23} \times 2 \times 1} \textrm{ g Ca}^{2+} = 3.2279641 \textrm{ g Ca}^{2+} (calculator answer)
= 3.2 g Ca ^{2+} (correct answer)
