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Chapter 17: Problem 67

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Short Answer

Expert verified
(a) No, Ca(OH)₂ will not precipitate from the solution because Q < Ksp (\(5 \times 10^{-14} < 6.5 \times 10^{-6}\)). (b) Yes, Ag₂SO₄ will precipitate when the solutions mix because Q > Ksp (\(4.24 \times 10^{-7} > 1.12 \times 10^{-5}\)).

Step by step solution

01

Find the concentration of OH-

The pH value is 8.0, so we can calculate the [OH-] concentration using the relationship: \(pOH = 14 - pH\) The pOH is equal to 6, so we can find the concentration of hydroxide ions (OH⁻) as follows: \[ [OH⁻] = 10^{-pOH} = 10^{-6} = 1 \times 10^{-6} \mathrm{M} \]
02

Calculate Ion Product (Q) and compare with Ksp

Now we will use the concentration of Ca²⁺ and OH⁻ ions to calculate the ion product (Q) for Ca(OH)₂. Since the given solution is 0.050 M CaCl₂, we have a [Ca²⁺] of 0.050 M. Then, the ion product (Q) for Ca(OH)₂ is: \[Q = [Ca²⁺][OH⁻]^2 = (0.050)(1 \times 10^{-6})^2 \] \[Q = 5 \times 10^{-14} \] The Ksp for Ca(OH)₂ is approximately \(6.5 \times 10^{-6}\). Since Q < Ksp, the system is not supersaturated, and the solution is not concentrated enough for precipitation to occur. (a) Answer: No, Ca(OH)₂ will not precipitate from the solution. (b) Determine if \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) will precipitate
03

Calculate final concentrations of ions after mixing

We're given 100 mL of 0.050 M AgNO₃ and 10 mL of 5.0 x 10⁻² M Na₂SO₄ When these solutions mix, the total volume becomes 110 mL, so we can calculate the final concentrations of the ions using the dilution formula: \[C_{1}V_{1} = C_{2}V_{2}\] For Ag⁺: \[(0.050)(100) = C_{Ag}(110)\] \[C_{Ag} = \frac{0.050 \times 100}{110} = 4.55 \times 10^{-2} M\] For SO₄²⁻: \[(5.0 \times 10^{-2})(10) = C_{SO4}(110)\] \[C_{SO4} = \frac{5.0 \times 10^{-2} \times 10}{110} = 4.55 \times 10^{-3} M\]
04

Calculate Ion Product (Q) and compare with Ksp

We will now use the final concentrations of Ag⁺ and SO₄²⁻ to calculate the ion product(Q) for Ag₂SO₄: \[Q = [Ag⁺]^2[SO₄^{2-}] = (4.55 \times 10^{-2})^2(4.55 \times 10^{-3})\] \[Q \approx 4.24 \times 10^{-7}\] The Ksp for Ag₂SO₄ is approximately \(1.12 \times 10^{-5}\). Since Q > Ksp, the solution is supersaturated, and Ag₂SO₄ will precipitate. (b) Answer: Yes, Ag₂SO₄ will precipitate when the solutions mix.

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Most popular questions from this chapter

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH},\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3},(\mathrm{c})\) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2+}\) in \(1.0 \mathrm{~L}\) of a solution that contains a total of \(1 \times 10^{-3} \mathrm{~mol}\) of nickel(II) ion and that is \(0.20 \mathrm{M}\) in \(\mathrm{NH}_{3}\)

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of \(\mathrm{pH} 7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood \(\mathrm{pH}\) is \(7.1 ?\)

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Calculate the molar solubility of \(\mathrm{Ni}(\mathrm{OH})_{2}\) when buffered at \(\mathrm{pH}(\) a) \(8.0,(\mathbf{b}) 10.0,(\mathrm{c}) 12.0\)
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