Solutions to Homework Assignment 11 CHM 152 Spring 2002

Transcription

1 Solutions to Homework Assignment 11 CHM 152 Spring Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid. Moles acid = 16.4 ml 1 L mol = mol 1000 ml 1 L Molar mass = g mol = 202 g/mol The neutralization reaction is: H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write: mol H2SO4 2molNaOH mol NaOH = ( L H 2 SO 4 ) 1Lsoln 1molH2SO4 = mol NaOH concentration of NaOH = mol NaOH 1000 ml = 0.25 M 50.0 ml soln 1 L HCOOH + Ba(OH) 2 (HCOO) 2 Ba + 2H 2 O Number of moles of HCOOH reacted = mol 1L 3 ( L) = mol HCOOH The mole ratio between Ba(OH) 2 and HCOOH is 1:2. Therefore, the molarity of the Ba(OH) 2 solution is: mol HCOOH 1molBa(OH) 2 1 2molHCOOH L = M (a) HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and phenolphthalein. (c) HCl is a strong acid and KOH is a strong base. Suitable indicators are all those listed with the exceptions of thymol blue, bromophenol blue, and methyl orange. HNO 3 is a strong acid and CH 3 NH 2 is a weak base. Suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue The weak acid equilibrium is HIn (aq) H + (aq) + In (aq)

2 We can write a K a expression for this equilibrium. Rearranging, + ] K a = [ H ][ In [ HIn] [ HIn] [ In ] = + [ H ] K a From the ph, we can calculate the H + concentration. [H + ] = 10 ph = 10 4 = M [HIn] [In ] + 4 [ H ] = = K a = 100 Since the concentration of HIn is 100 times greater than the concentration of In, the color of the solution will be that of HIn, the nonionized formed. The color of the solution will be red Solubility is the number of grams of BaSO 4 that will dissolve per liter of solution. Molar solubility is the number of moles of BaSO 4 that will dissolve per liter of solution. The solubility product is: K sp = [Ba 2+ ][SO 4 2 ] The K sp values are very large indicating that there are almost exclusively products in solution at equilibrium. We assume that soluble ionic compounds completely dissociate into ions. Therefore, the equilibrium constant is not needed (a) The solubility equilibrium is given by the equation AgI(s) Ag + (aq) + I (aq) The expression for K sp is given by K sp = [Ag + ][I ] The value of K sp can be found in Table 16.2 of the text. If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be [I K 17 sp ] = = = M 9 [ Ag ] The value of K sp for aluminum hydroxide can be found in Table 16.2 of the text. The equilibrium expressions are: Al(OH) 3 (s) Al 3+ (aq) + 3OH (aq) K sp = [Al 3+ ][OH ] 3

3 Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is: 3+ K 33 sp [Al ] = = = [ OH ] 3 M 9 3 ( ) What is the ph of this solution? Will the aluminum concentration change if the ph is altered? The charges of the M and X ions are +3 and 2, respectively (are other values possible?). We first calculate the number of moles of M 2 X 3 that dissolve in 1.0 L of water 17 1mol 19 Mo les M2X 3 = ( g) = mol 288 g The molar solubility, s, of the compound is therefore M. At equilibrium the concentration of M 3+ must be 2s and that of X 2 must be 3s. (See Table 16.3 of the text.) K sp = [M 3+ ] 2 [X 2 ] 3 = [2s] 2 [3s] 3 = 108s 5 Since these are equilibrium concentrations, the value of K sp can be found by simple substitution K sp = 108s 5 = 108( ) 5 = Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product expression. CaF 2 (s) Ca 2+ (aq) + 2 F (aq) K sp = [Ca 2+ ][F ] 2 Step 2: A certain amount of calcium fluoride will dissociate in solution. Let s represent this amount as s. Since one unit of CaF 2 yields one Ca 2+ ion and two F ions, at equilibrium [Ca 2+ ] is s and [F ] is 2s. We summarize the changes in concentration as follows: CaF 2 (s) Ca 2+ (aq) + 2 F (aq) Initial (M): 0 0 Change (M): s +s +2s Equilibrium (M): s 2s Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression. Therefore, the concentration of CaF 2 is not important. Step 3: Substitute the value of K sp and the concentrations of Ca 2+ and F in terms of s into the solubility product expression to solve for s, the molar solubility. K sp = [Ca 2+ ][F ] = (s)(2s) = 4s 3 s = molar solubility = mol/l The molar solubility indicates that mol of CaF 2 will dissolve in 1 L of an aqueous solution.

4 16.48 First we can calculate the OH concentration from the ph. poh = ph poh = = 4.32 [OH ] = 10 poh = = M The equilibrium equation is: MOH (s) M + (aq) + OH (aq) From the balanced equation we know that [M + ] = [OH ] K sp = [M + ][OH ] = ( ) 2 = (a) Set up a table to find the equilibrium concentrations in pure water. Pb 2+ (aq) + 2 Br (aq) Initial (M) 0 0 Change (M) s +s +2s Equilibrium (M) s 2s K sp = [Pb 2+ ][Br ] = (s)(2s) 2 s = molar solubility = M Set up a table to find the equilibrium concentrations in 0.20 M KBr. KBr is a soluble salt that ionizes completely giving a initial concentration of Br = 0.20 M. Pb 2+ (aq) + 2 Br (aq) Initial (M) Change (M) s +s +2s Equilibrium (M) s s K sp = [Pb 2+ ][Br ] = (s)( s) (s)(0.20) 2 s = molar solubility = M Thus, the molar solubility of PbBr 2 is reduced from M to M as a result of the common ion (Br ) effect. (c) Set up a table to find the equilibrium concentrations in 0.20 M Pb(NO 3 ) 2. Pb(NO 3 ) 2 is a soluble salt that dissociates completely giving an initial concentration of [Pb 2+ ] = 0.20 M. Pb 2+ (aq) + 2 Br (aq) Initial (M): Change (M): s +s +2s Equilibrium (M): s 2s

5 K sp = [Pb 2+ ][Br ] = ( s)(2s) (0.20)(2s) 2 s = molar solubility = M Thus, the molar solubility of PbBr 2 is reduced from M to M as a result of the common ion (Pb 2+ ) effect. Additional Problem: 10.0 ml of 0.25 M HBr is titrated with 0.10 M KOH. Calculate the ph at the following points in the titration. a ml of KOH added ph = 0.60 b ml of KOH added ph = 1.12 c ml of KOH added ph = 1.78 d ml of KOH added ph = 7.00 (equivalence point) e ml of KOH added ph = 12.35