U3.8 · Representations of SolutionsPractice · 溶液的粒子图表示

$\ce{Pb^2+ + 2I- -> PbI2(s)}$. The 4 $\ce{I-}$ ions can precipitate only 2 $\ce{Pb^2+}$, forming 2 $\ce{PbI2}$ and leaving 1 $\ce{Pb^2+}$. $\ce{K+}$ (4) and $\ce{NO3-}$ (6) are spectator ions and all stay in solution. A has no $\ce{Pb^2+}$ left, so its charges don’t balance; B shows no reaction; D has 4 extra I, so atoms aren’t conserved.

The 3 $\ce{Cl-}$ ions come from 3 NaCl, which bring 3 $\ce{Na+}$. There are as many $\ce{Na2CO3}$ as NaCl, so 3, each giving 2 $\ce{Na+}$, for 6 more. Total $\ce{Na+}$ = 3 + 6 = 9, along with 3 $\ce{CO3^2-}$ ions. Check the charge: 9 − 3 − 2 × 3 = 0.

Water is polar: O carries a partial negative charge and the H atoms carry partial positive charges. The cation $\ce{K+}$ attracts the δ− oxygen end, and the anion $\ce{F-}$ attracts the δ+ hydrogen end. These are ion–dipole attractions.

In 0.30 M NaCl, $[\ce{Na+}]$ = $[\ce{Cl-}]$ = 0.30 M, drawn as 6 ions each, so each ion stands for 0.05 M. In 0.10 M $\ce{MgCl2}$, $[\ce{Mg^2+}]$ = 0.10 M and $[\ce{Cl-}]$ = 0.20 M: 2 $\ce{Mg^2+}$ and 4 $\ce{Cl-}$. B’s charges don’t balance, C is 0.30 M $\ce{MgCl2}$, and D is 0.050 M.

The diagram has 2 $\ce{Al^3+}$ and 6 $\ce{NO3-}$ (a 1 : 3 ratio, so the charges balance). n($\ce{NO3-}$) = 6 × 0.020 = 0.120 mol, so $[\ce{NO3-}]$ = 0.120 ÷ 0.500 = 0.240 M. $[\ce{Al^3+}]$ is a third of that, 0.0800 M.