If 20.0 $\mathrm{cm^3}$ of 0.050 M $\mathrm{Ba(NO_3)_2}$ are mixed with 20.0 $\mathrm{cm^3}$ of 0.020 M NaF, will $\mathrm{BaF_2}$ precipitate ? $K_{sp}$ of $\mathrm{BaF_2}$ is $1.7 imes 10^{-6}$ at 298 K.

Question

If 20.0 $\mathrm{cm^3}$ of 0.050 M $\mathrm{Ba(NO_3)_2}$ are mixed with 20.0 $\mathrm{cm^3}$ of 0.020 M NaF, will $\mathrm{BaF_2}$ precipitate ? $K_{sp}$ of $\mathrm{BaF_2}$ is $1.7 	imes 10^{-6}$ at 298 K.

Accepted Answer

> [!TLDR]
> After mixing, $[\mathrm{Ba^{2+}}] = 0.025$ M and $[\mathrm{F^-}] = 0.010$ M; IP $= 2.5	imes10^{-6}$ just exceeds $K_{sp} = 1.7	imes10^{-6}$, so precipitation occurs.

**Step 1.** Total volume after mixing $= 20.0 + 20.0 = 40.0$ cm$^3$, so each solute is diluted by the factor $20/40$: $[\mathrm{Ba(NO_3)_2}] = 0.050	imes\dfrac{20}{40} = 0.025$ M and $[\mathrm{NaF}] = 0.020	imes\dfrac{20}{40} = 0.010$ M.

**Step 2.** Both salts are strong electrolytes, so $[\mathrm{Ba^{2+}}] = 0.025$ M and $[\mathrm{F^-}] = 0.010$ M in the mixed solution.

**Step 3.** $\mathrm{BaF_2}$ dissolves as $\mathrm{BaF_2(s) ightleftharpoons Ba^{2+}(aq) + 2F^-(aq)}$, so its ionic product is $	ext{IP} = [\mathrm{Ba^{2+}}][\mathrm{F^-}]^2 = 0.025	imes(0.01)^2 = 2.5	imes10^{-6}$. …

Q.If 20.0 $\mathrm{cm^3}$ of 0.050 M $\mathrm{Ba(NO_3)_2}$ are mixed with 20.0 $\mathrm{cm^3}$ of 0.020 M NaF, will $\mathrm{BaF_2}$ precipitate ? $K_{sp}$ of $\mathrm{BaF_2}$ is $1.7 \times 10^{-6}$ at 298 K.

Concept understanding — Solubility Product (Ksp)

Q.If 20.0 cm3\mathrm{cm^3}cm3 of 0.050 M Ba(NO3)2\mathrm{Ba(NO_3)_2}Ba(NO3​)2​ are mixed with 20.0 cm3\mathrm{cm^3}cm3 of 0.020 M NaF, will BaF2\mathrm{BaF_2}BaF2​ precipitate ? KspK_{sp}Ksp​ of BaF2\mathrm{BaF_2}BaF2​ is 1.7×10−61.7 \times 10^{-6}1.7×10−6 at 298 K.

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Q.If 20.0 $\mathrm{cm^3}$ of 0.050 M $\mathrm{Ba(NO_3)_2}$ are mixed with 20.0 $\mathrm{cm^3}$ of 0.020 M NaF, will $\mathrm{BaF_2}$ precipitate ? $K_{sp}$ of $\mathrm{BaF_2}$ is $1.7 \times 10^{-6}$ at 298 K.

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