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Chapter 17: Problem 72

Suppose that a 10 -mL sample of a solution is to be tested for \(\mathrm{I}^{-}\) ion by addition of 1 drop \((0.2 \mathrm{~mL})\) of $0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ What is the minimum number of grams of \(\mathrm{I}^{-}\) that must be present for \(\mathrm{PbI}_{2}(s)\) to form?

Short Answer

Expert verified
The minimum number of grams of $\mathrm{I}^{-}$ ions needed for $\mathrm{PbI}_{2}(s)$ to form when 0.2 mL of a 0.10 M $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ solution is added to a 10-mL sample is approximately \(5.80 \times 10^{-4}\) grams.

Step by step solution

01

Find the solubility product constant for PbI₂

From the given information, we have to find the solubility product constant (Ksp) of lead(II) iodide (PbI₂). With the help of a reference table or other data source, we find that the Ksp value for PbI₂ is 7.1 x 10⁻₉.
02

Write the balanced equation for PbI₂(s) dissolution

The dissolution of PbI₂(s) in water can be represented as follows: \[PbI_{2}(s) \leftrightarrow Pb^{2+}(aq) + 2I^-(aq)\]
03

Determine the initial concentrations

Since 0.2 mL of 0.10 M Pb(NO₃)₂ is added to the solution, we can convert the volume to liters and then determine the moles of Pb²⁺ ions introduced: Volume of Pb(NO₃)₂ = \(0.2\,mL \times \frac{1\,L}{1000\, mL} = 2 × 10^{-4} L\) Moles of Pb²⁺ ions = Molarity × Volume = \(0.10\,M \times 2 × 10^{-4} L = 2 × 10^{-5}\,mol\) Since the volume of the sample solution is 10 mL, the final volume of the mixture after adding Pb(NO₃)₂ is 10.2 mL. Hence, the initial concentration of Pb²⁺ ions in the mixture after addition is: Concentration of Pb²⁺ ions = \(\frac{2 × 10^{-5}\, mol}{10.2 × 10^{-3} L} = 1.96 × 10^{-3} M\) Considering the 1:2 stoichiometry in the chemical equation, Pb²⁺ and I⁻ have a 1:2 molar ratio.
04

Evaluate the minimum concentration of I⁻ ions required using Ksp

Let x be the minimum concentration of I⁻ ions needed. The solubility product expression for the dissociation of PbI₂ can be written as follows: \[Ksp = [Pb^{2+}][I^-]^2 = (1.96 × 10^{-3} - x)(2x)^2\] Since the solubility product constant Ksp is given as 7.1 x 10⁻₉, we substitute this value in the equation and solve for x: \[7.1 \times 10^{-9} = (1.96 × 10^{-3} - x)(2x)^2\]
05

Solve for x, the minimum concentration of I⁻ ions

By solving this equation for x, we can find the minimum concentration of I⁻ ions for PbI₂ to form. The equation will be challenging to solve algebraically, so we can use an iterative method or graphing calculator to find the roots. Upon solving, we get x ≈ \(4.57 × 10^{-4} M\) as the minimum concentration of I⁻ ions needed.
06

Calculate the minimum grams of I⁻ ions present in the 10-mL sample

Now that we have the minimum concentration of I⁻ ions, we can determine the minimum grams of I⁻ ions present in the 10-mL sample. First, convert the concentration to moles of I⁻ ions in a 10-mL sample: Moles of I⁻ ions = Concentration × Volume = \(4.57 × 10^{-4} M \times 10 × 10^{-3} L = 4.57 × 10^{-6}\,mol\) Next, convert moles to grams using the molar mass of iodide ions (I⁻ = 127 g/mol): Minimum grams of I⁻ ions = Moles × Molar Mass = \(4.57 × 10^{-6} mol \times 127 \frac{g}{mol} = 5.80 × 10^{-4} g\) Thus, the minimum number of grams of I⁻ ions needed for PbI₂ to form is approximately \(5.80 × 10^{-4} g\).

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Most popular questions from this chapter

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of $1.00 \mathrm{MHCl}$.

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(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate $K_{s p} .(\mathbf{b})\( If \)0.0490 \mathrm{~g}\( of \)\mathrm{AgIO}_{3}$ dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(\mathrm{K}_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and $K_{f}=1.1 \times 10^{5}\( for \)\mathrm{AgCl}_{2}^{-},$ calculate the equilibrium constant for the following reaction: $$ \mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) $$
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