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57. A solution contains 1.0×10−5MNa3PC41.0 \times 10^{-5} \mathrm{MNa}_{3} \mathrm{PC}_{4}. W...
Jan 29, 2024
57. A solution contains 1.0×10−5MNa3PC41.0 \times 10^{-5} \mathrm{MNa}_{3} \mathrm{PC}_{4}. What is the minimum concentration of AgNO3\mathrm{AgNO}_{3} that would cause precipitation of solid Ag3PO4(Ksp=1.8×10−18)\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?
Solution by Steps
step 1
To find the minimum concentration of AgNO3\mathrm{AgNO}_{3} that would cause precipitation, we use the solubility product constant KspK_{\mathrm{sp}}
step 2
The KspK_{\mathrm{sp}} for Ag3PO4\mathrm{Ag}_{3}\mathrm{PO}_{4} is given as 1.8×10−181.8 \times 10^{-18}. The dissociation of Ag3PO4\mathrm{Ag}_{3}\mathrm{PO}_{4} in water is represented by the equation: Ag3PO4(s)⇌3Ag+(aq)+PO43−(aq)\mathrm{Ag}_{3}\mathrm{PO}_{4}(s) \rightleftharpoons 3\mathrm{Ag}^{+}(aq) + \mathrm{PO}_{4}^{3-}(aq)
step 3
The KspK_{\mathrm{sp}} expression is Ksp=[Ag+]3[PO43−]K_{\mathrm{sp}} = [\mathrm{Ag}^{+}]^{3}[\mathrm{PO}_{4}^{3-}]. Given that the concentration of PO43−\mathrm{PO}_{4}^{3-} from Na3PC4\mathrm{Na}_{3}\mathrm{PC}_{4} is 1.0×10−51.0 \times 10^{-5} M, we can substitute this into the KspK_{\mathrm{sp}} expression
step 4
Substituting the known values into the KspK_{\mathrm{sp}} expression gives us 1.8×10−18=[Ag+]3(1.0×10−5)1.8 \times 10^{-18} = [\mathrm{Ag}^{+}]^{3}(1.0 \times 10^{-5})
step 5
To solve for [Ag+][\mathrm{Ag}^{+}], we divide both sides by 1.0×10−51.0 \times 10^{-5} to get [Ag+]3=1.8×10−181.0×10−5[\mathrm{Ag}^{+}]^{3} = \frac{1.8 \times 10^{-18}}{1.0 \times 10^{-5}}
step 6
Calculating the right-hand side gives [Ag+]3=1.8×10−13[\mathrm{Ag}^{+}]^{3} = 1.8 \times 10^{-13}
step 7
Taking the cube root of both sides to solve for [Ag+][\mathrm{Ag}^{+}] gives [Ag+]=1.8×10−133[\mathrm{Ag}^{+}] = \sqrt[3]{1.8 \times 10^{-13}}
step 8
Using a calculator, we find that [Ag+]≈5.65×10−5[\mathrm{Ag}^{+}] \approx 5.65 \times 10^{-5} M
Answer
The minimum concentration of AgNO3\mathrm{AgNO}_{3} that would cause precipitation is approximately 5.65×10−55.65 \times 10^{-5} M.
Key Concept
Solubility Product Constant (Ksp)
Explanation
The solubility product constant is used to determine the solubility of a compound in a solution. It is the product of the molar concentrations of the ions each raised to the power of its coefficient in the dissociation equation.
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