/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 (a) Will \(\mathrm{Ca}(\mathrm{O... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 63

(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) Upon adjusting the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) to \(8.0\), \(\mathrm{Ca}(\mathrm{OH})_{2}\) will not precipitate since the ion product, Q = \(2.5 \times 10^{-13}\), is less than the solubility product constant, Ksp = \(4.68 \times 10^{-11}\). (b) 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, \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) will not precipitate since the ion product, Q = \(9.3 \times 10^{-6}\), is less than the solubility product constant, Ksp = \(1.2 \times 10^{-5}\).

Step by step solution

01

Calculate the concentration of hydroxide ions

Given the pH of the solution is \(8.0\), we can find the hydroxide ion concentration (\([\mathrm{OH}^{-}]\)) using the following relationship: \(pH + pOH = 14\) Since \(pH = 8.0\), \(pOH = 14 - pH = 6\) \( [\mathrm{OH}^{-}] = 10^{-pOH} = 10^{-6} \, \mathrm{M} \)
02

Calculate the ion-product for Calcium Hydroxide

The ion-product for Calcium Hydroxide can be calculated as follows: Q = [Ca虏鈦篯[OH鈦籡虏 Since the calcium ion concentration is given as \(0.050 \, \mathrm{M}\), we can plug the values in the equation: Q = (0.050)(10^{-6})虏 Q = \(2.5 \times 10^{-13}\)
03

Compare Q to Ksp for Calcium Hydroxide

The solubility product constant (Ksp) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(4.68 \times 10^{-11}\). Since Q (2.5 x 10^(-13)) is less than Ksp (4.68 x 10^(-11)): Q < Ksp Therefore, \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitation will not occur in this solution. (b)
04

Calculate the concentrations of ions in the mixed solution

In this case, we have to find the \(\mathrm{Ag}^+\) and \(\mathrm{SO}_4^{2-}\) ion concentrations in the mixed solution. Volume of the mixed solution = \(100\,\mathrm{mL} + 10\,\mathrm{mL} = 110\,\mathrm{mL} \) Silver ion concentration: C1 = 0.050 M (initial concentration in \(\mathrm{AgNO}_{3}\) solution) V1 = 100 mL (volume of \(\mathrm{AgNO}_{3}\) solution) V2 = 110 mL (total volume) C2 = (C1 脳 V1) / V2 C2 = (0.050 脳 100) / 110 C2 = 0.0455 M (concentration of \(\mathrm{Ag}^+\) ions in the mixed solution) Sulfate ion concentration: C1 = \(5.0 \times 10^{-2} \mathrm{M}\) (initial concentration in \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution) V1 = 10 mL (volume of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) solution) V2 = 110 mL (total volume) C2 = (C1 脳 V1) / V2 C2 = (\(5.0 \times 10^{-2}\) 脳 10) / 110 C2 = \(4.55 \times 10^{-3} \,\mathrm{M}\) (concentration of \(\mathrm{SO}_4^{2-}\) ions in the mixed solution)
05

Calculate the ion-product for Silver Sulfate

The ion-product for Silver Sulfate can be calculated as follows: Q = [Ag鈦篯虏[SO鈧劼测伝] Plugging in the values from the calculation above: Q = (0.0455)虏(\(4.55 \times 10^{-3}\)) Q = \(9.3 \times 10^{-6}\)
06

Compare Q to Ksp for Silver Sulfate

The solubility product constant (Ksp) for \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) is \(1.2 \times 10^{-5}\). Since Q (9.3 x 10^(-6)) is less than Ksp (1.2 x 10^(-5)): Q < Ksp Therefore, \(\mathrm{Ag}_{2}\mathrm{SO}_{4}\) precipitation will not occur when the solutions are mixed.

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Most popular questions from this chapter

(a) Calculate the percentionization of \(0.085 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .\) (b) Calculate the percent ionization of \(0.095 \mathrm{M}\) lactic acid in a solution containing \(0.0075 \mathrm{M}\) sodium lactate.

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Suppose you want to do a physiological experiment that calls for a pH \(6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) 6.50? (lgnore any volume change.)

How does titration of a strong, monoprotic acid with a strong base differ from titration of a weak, monoprotic acid with a strong base with respect to the following: (a) quantity of base required to reach the equivalence point, (b) \(\mathrm{pH}\) at the beginning of the titration, \((\mathrm{c}) \mathrm{pH}\) at the equivalence point, (d) \(\mathrm{pH}\) after addition of a slight excess of base, (e) choice of indicator for determining the equivalence point?
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