/*! 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 107 The value of \(K_{s p}\) for \(\... [FREE SOLUTION] | 91影视

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The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20} .\) The \(\mathrm{AsO}_{4}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{a 1}=\right.\) \(\left.2.22 ; \mathrm{p} K_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right)\) (a) Calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water. (b) Calculate the pH of a saturated solution of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water.

Short Answer

Expert verified
(a) The molar solubility of Mg鈧(AsO鈧)鈧 in water is \(7.34 \times 10^{-3}\) M. (b) The pH of a saturated solution of Mg鈧(AsO鈧)鈧 in water is approximately 2.22.

Step by step solution

01

(Establish the equilibrium)

Write the dissociation equilibrium of Mg3(AsO4)2 and establish an ICE table. Mg鈧(AsO鈧)鈧 (s) 鈬 3Mg虏鈦 (aq) + 2AsO鈧劼斥伝 (aq) Initial concentrations: [Mg虏鈦篯 = 0 [AsO鈧劼斥伝] = 0 Changes: [Mg虏鈦篯 = +3s [AsO鈧劼斥伝] = +2s Final concentrations after dissociation: [Mg虏鈦篯 = 3s [AsO鈧劼斥伝] = 2s
02

(Use the Ksp to find solubility)

Use the given Ksp to establish the solubility equation and solve for the solubility 's' of Mg3(AsO4)2. Ksp = [Mg虏鈦篯鲁[AsO鈧劼斥伝]虏 2.1 脳 10鈦宦测伆 = (3s)鲁(2s)虏
03

(Find the molar solubility of Mg3(AsO4)2)

Solve for the solubility 's' of Mg3(AsO4)2. s^9 = (2.1 脳 10鈦宦测伆) / (3鲁 脳 2虏) s^9 = 2.59 脳 10鈦宦猜 s = 0.00734 (as ninth root of s^9) Molar solubility of Mg鈧(AsO鈧)鈧 = 7.34 脳 10鈦宦 M.
04

(Find the concentration of ions in the saturated solution)

With the molar solubility found, calculate the concentration of ions in the saturated solution. [Mg虏鈦篯 = 3s = 3(7.34 脳 10鈦宦) = 2.20 脳 10鈦宦 M [AsO鈧劼斥伝] = 2s = 2(7.34 脳 10鈦宦) = 1.47 脳 10鈦宦 M
05

(Calculate the pH of the saturated solution)

Use the pKa values given and ion concentrations to determine the pH of the saturated solution. Considering the first proton dissociation in H鈧傾sO鈧: H鈧傾sO鈧 鈬 H鈦 + H鈧侫sO鈧勨伝 pKa1 = 2.22 Ka1 = [H鈦篯[H鈧侫sO鈧勨伝] / [H鈧傾sO鈧刔 [H鈦篯 = 10^(-pKa1) = 10^(-2.22) = 6.027 脳 10鈦宦 M pH = -log[H鈦篯 = -log(6.027 脳 10鈦宦) = 2.22 (approximately) (a) The molar solubility of Mg鈧(AsO鈧)鈧 in water is 7.34 脳 10鈦宦 M. (b) The pH of a saturated solution of Mg鈧(AsO鈧)鈧 in water is approximately 2.22.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ksp (Solubility Product Constant)
The solubility product constant, often abbreviated as \(K_{sp}\), is a crucial concept in understanding the solubility of sparingly soluble ionic compounds in water. It represents the equilibrium constant for the dissolution reaction of a solid substance. For our case, the dissolution of \(\text{Mg}_3(\text{AsO}_4)_2\) in water establishes itself in the following equilibrium:\[\text{Mg}_3(\text{AsO}_4)_2 (s) \rightleftharpoons 3\text{Mg}^{2+} (aq) + 2\text{AsO}_4^{3-} (aq)\]The value of \(K_{sp}\) is expressed as:\[K_{sp} = [\text{Mg}^{2+}]^3[\text{AsO}_4^{3-}]^2\]In the equation above, \([\text{Mg}^{2+}]\) and \([\text{AsO}_4^{3-}]\) are the molar concentrations of the ions in the solution at equilibrium. For this particular compound, \(K_{sp}\) is given as \(2.1 \times 10^{-20}\), which is quite small, indicating very low solubility in water. To calculate molar solubility, set up an equation based on how each ion's concentration changes proportionately with the dissolution of \(\text{Mg}_3(\text{AsO}_4)_2\). By determining the solubility \(s\), you find how many moles of the solid will dissolve per liter of water. Here, \(s\) was found to be \(7.34 \times 10^{-3}\) M.
Ionic Equilibrium
Ionic equilibrium deals with the state of balance between the ions in a solution when a particular compound has dissolved partially. At equilibrium, the rate of dissociation (dissolving into ions) matches the rate of precipitation (ions combining back into a solid).**Establishing Equilibrium**:For \(\text{Mg}_3(\text{AsO}_4)_2\), as it dissolves, there will be initial concentrations of ions as zero:- \([\text{Mg}^{2+}] = 0\)- \([\text{AsO}_4^{3-}] = 0\)As the equilibrium establishes, these concentrations increase:- \([\text{Mg}^{2+}] = 3s\)- \([\text{AsO}_4^{3-}] = 2s\)The equilibrium concentrations directly depend on the molar solubility \(s\), indicating how much of the salt dissolves in the solution. Use these relationships to plug into the \(K_{sp}\) equation, which emphasizes its role in buffering between dissociation and reformation of the solid under non-ideal, low solubility conditions.
Acid Dissociation Constant (pKa)
The acid dissociation constant \((pK_a)\) is instrumental in understanding how weak acids behave in solution, affecting both solubility and pH of the solution. In the given problem, we analyzed arsenic acid \(\text{H}_3\text{AsO}_4\), which can lose protons in three steps, each characterized by its own \(pK_a\):- \(pK_{a1} = 2.22\)- \(pK_{a2} = 6.98\)- \(pK_{a3} = 11.50\)These values indicate the strength and the degree to which \(\text{H}_3\text{AsO}_4\) loses its protons in solution. The first dissociation has the lowest \(pK_a\), which makes \(\text{H}_3\text{AsO}_4\) a stronger acid compared to its other forms. **pH of the Solution**:In a saturated solution of \(\text{Mg}_3(\text{AsO}_4)_2\), where \(\text{AsO}_4^{3-}\) ions are present, using the \(pK_{a1}\), we can calculate the concentration of \([H^+]\). This initial concentration helps in determining the pH of the solution using the formula:\[pH = -\log[\text{H}^+]\]Here, the calculated \([H^+]\) of \(6.027 \times 10^{-3}\) M provides a pH of approximately 2.22, showing the acidic nature of the solution. Understanding \(pK_a\) values can also inform us about the polyprotic nature of acids and their sequential ionization in different pH environments, influencing the solubility equilibrium.

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Most popular questions from this chapter

Consider the equilibrium $$ \mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

(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 molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) 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.

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

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-},\) helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) per \(0.500 \mathrm{~L}\) of solution?

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},(\mathbf{b}) \mathrm{NH}_{3}\) titrated with \(\mathrm{HCl},\) (c) KOH titrated with HBr.

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