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Determine the solubilities of silver chr...

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at `298K` from theor solubility product constants given in Table `7.9`. Determine also the molarities of individual ions.

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To determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride, and mercurous iodide at 298 K, we will use their solubility product constants (Ksp) provided in the problem statement. We will also calculate the molarities of the individual ions produced upon dissolution. ### Step-by-Step Solution: 1. **Silver Chromate (Ag2CrO4)**: - Dissociation: \[ \text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \] - Let the solubility be \( s \). Then: ...
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Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K form their solubility product constants given below. Determine also the molarities of individual ions. K_(SP(Ag_(2)CrO_(4)))=1.1xx10^(-12) , K_(SP(BaCrO_(4)))=1.2xx10^(-10) , K_(SP[Fe(OH)_(3)])=1.0xx10^(-38) , K_(SP(PbCI_(2)))=1.6xx10^(-5) , K_(SP(Hg_(2)I_(2)))=4.5xx10^(-29) .

Determine the solubility of silver chromate at 298 K given its K_(sp) value is 1.1xx10^(-12) ?

(a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride. (b) Given are the conductivity and molar conductivity of NaCI solutions at 298 K at different concentrations : Compare the variation of conductivity and molar conductivity of NaCI solutions on dilution. Give reason. (c) 0.1 M KCI solution offered a resistance of 100 ohms in conductivity cell at 298 K. If the cell constant of the cell is 1.29 cm^(-1) , calculate the molar conductivity of KCI solution.

A solution which remains in equilibrium with undissolved solute , in contact , is said to be saturated . The concentration of a saturated solution at a given temperature is a called solubility . The product of concentration of ions in a saturated solution of an electrolyte at a given temperature is called solubility product (K_(sp)) . For the electrolyte A_(x),B_(y) with solubility S. The solubility product (K_(sp)) is given as K_(sp) = x^(x) xx y^(y) xx S^(x-y) . While calculating the solubility of a sparingly . soluable salt in the presence of some strong electrolyte containing a common ion , the common ion concentration is practically equal to that of strong electrolyte containing a common ion . the common ion soncentration is practically equal to that of strong electrolyte . If in a solution , the ionic product of an electroylte exceeds its K_(sp) value at a particular temperature , then precipitation occurs . If two or more electrolyte are presentt in the solution , then by the addition of some suitable reagent , precipitation generally occurs in increasing order of their k_(sp) values . Solubility of some sparingly soluable salts , is sometimes enhanced through complexation . While we are calculating the solubility of some sparingly or pH of an electrolyte , the nature of cation of anion should be checked carefully whether there ion (s) are capable of undergoing hydrolysis or not . If either or both of the ions are capable of undergoing hydrolysis , it should be taken into account in calculating the solubility . While calculating the pH of an amphiprotic species , it should be checked whether or not cation can undergo hydrolysis . Total a_(H^(-)) = sqrt(K_(a_(1)xxK_(a_(2)))) (if cation do not undergo hydrolysis ) a_(H^(+)) = sqrt(K_(a_(1))((K_(w))/(K_(b)) - K_(a_(2)))) (if cation also undergoes hydrolysis ) where symbols have usual meaning . Solubility of solids into liquids is a function of temperature alone but solubility of gases into liquids is a function of temperature as well as pressure . The effect of pressure on solubility of gases into liquids is governed by Henry's law . The solubility of BaSO_(4) in 0.1 M BaCl_(2) solution is (K_(sp) " of " BaSO_(4) = 1.5 xx 10^(-9))

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Determine the molar solubility of MgF_(2) from its solubility product K_(sp)=4xx10^(-9) :

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from The soluvility molar solubility of ferric hydroxide in aqueous solution is 6xx10^(-38) at 298 K. the solubility of Fe^(3+) ion will increase when the :

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Will a precipitate from if 1 volume of 0.1 volume of 0.1 MPb^(2+) ion solution in mixed with 3 volume of 0.3 M Cl^(-) ion solution ? ["Givem":K_(sp)(PbCl_(2))=1.7xx10^(-5)M^(3)]

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from At 25^(@)C, will a precipitate of Mg (OH)_(2) from when a 0.0001 M solution of Mg(NO_(3))_(2) is adjusted to a pH of 9.0 ? At what minimum value of pH will precipition start ? ["Given" : K_(sp)(Mg(OH)_(2))=10^(-11)M^(3)]

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Will a precipitate from if 50 cm^(3) of 0.01 M AgNO_(3) and 50 cm^(3) of 2xx10^(-5) M NaCl are mixed? ["Given": K_(sp)(AgCl)=10^(-10)M^(2)]

A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at 100.15^(@)C . To the resulting solution 0.2 mole NaCI was added. The resulting solution was found to freeze at 0.83^(@)C . Determine solubility product of PbCI_(2) . Given K_(b) = 0.5 and K_(f) = 1.86 . Assume molality to be equal to molarity in all case.

CENGAGE CHEMISTRY ENGLISH-NCERT BASED EXERCISE-Chemical Equilibrium
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  16. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorat...

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