Advanced Chemistry Practice Problems

Transcription

1 Finding ph 1. Question: Determine the ph for each of the given solutions. a M HNO3 b M CH3COOH, a = c M CHOOH, a = Answer: The method to determine the ph of a solution will depend on whether the acid is strong or weak. a M HNO3 Nitric acid is a strong, monoprotic acid. Therefore, [HNO3 = [H + = M ph log[h ph log[0.150 ph b M CH3COOH, a = For a weak acid, such as acetic acid, an equilibrium equation and ICE table must be used to determine the ph. Weak acids do not completely dissociate so we cannot assume that [H + = [CH3COOH. The a value (given in the problem) is CH3COOH(aq) CH3COO - (aq) + H + (aq) INITIAL: The initial concentration of CH3COOH is known to be M. Since the problem concerns the ionization of acetic acid into its ions, the initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.

2 CH3COOH (M) CH3COO - (M) H + (M) I M 0 0 C -x +x +x E x x x The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [H a -5 CH COO H 3 CH 3 x x x COOH The value of a is small, so the assumption that x << (much less) can be made. -5 x x Now, the value of x can be determined. x x The value of x is less than 5% of the value from which it was subtracted (0.150) so the assumption is valid. The equilibrium value of [H + = x, therefore [H + = The ph of the solution can now be calculated. ph log[h ph ph log( The answer is reasonable because the ph of the M HNO3 solution is lower (more acidic) than the ph of the M CH3COOH solution. This is expected because HNO3 is a strong acid and CH3COOH is a weak acid. c M CHOOH, a = Formic acid, similar to acetic acid, is a weak acid. Therefore, an equilibrium equation and ICE table must be used to determine the ph of the solution. Weak acids do not completely ionize so we cannot assume that [H + = [CHOOH. The a value (given in the problem) is )

3 CHOOH(aq) CHOO - (aq) + H + (aq) INITIAL: The initial concentration of CHOOH is known to be M. Since the problem concerns the dissociation of formic acid into its ions, the initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column. CHOOH (M) CHOO - (M) H + (M) I M 0 0 C -x +x +x E x x x The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [H a -4 CHOO H x x x CHOOH The value of a is small, so the assumption that x << can be made. -4 x x Now, the value of x can be determined. x x The value of x is 4.8% of the value from which it was subtracted (0.150) so the assumption is valid. The equilibrium value of [H + = x, therefore [H + = The ph of the solution can now be calculated. 5

4 ph log[h ph log( ph 2.14 The answer is reasonable because the ph of the M HNO3 solution is lower (more acidic) than the ph of the M CHOOH solution. This is expected because HNO3 is a strong acid and CHOOH is a weak acid. The ph of CHOOH is also less than CH3COOH because it is a strong acid which is evidence by the greater a value for CHOOH. 2. Question: A solution of M hypochlorous acid (HOCl) has a a = , what is the percent ionization of this solution? 3 ) Answer: Solving for the percent ionization of a weak acid begins by solving for the [H +. Once [H + is known, the percent ionization can be calculated. HOCl is a weak acid so we cannot assume that [H + = [HOCl. The a value (given in the problem) is HOCl(aq) OCl - (aq) + H + (aq) INITIAL: The initial concentration of HOCl is given as M. The initial concentrations of the ions are zero. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column. HOCl (M) OCl - (M) H + (M) I M 0 0 C -x +x +x E x x x The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [H +.

5 a -8 OCl H HOCl x x x The value of a is small, so the assumption that x << can be made. -8 x x Now, the value of x can be determined. x x The value of x is less than 5% of the value from which it was subtracted (0.100) so the assumption is valid. The goal of calculating percent ionization is to determine what fraction of the acid ionized (dissociated into ions). The formula for percent ionization is % ionization 9 H HOCl With the initial concentration of HOCl and the calculated concentration of H +, the percent ionization can be calculated. % ionization % ionization % The percent ionization can serve as another way to compare the strength of acids. The higher the percent ionization, the strong the acid. HOCl is a weak acid with a very low a value has a very low percent ionization.

6 Strong and Weak Bases 1. Question: Identify each of the following as a strong or weak base. a. OH b. NH3 c. CH3NH2 d. Ba(OH)2 Answer: Strong bases contain the hydroxide anion while most weak bases will be an ammine compound (contain an N as a central atom). a. OH strong b. NH3 weak c. CH3NH2 weak d. Ba(OH)2 strong 2. Question: Complete the table. Assume all solutions are at 25 C. Solution [H + ph [OH - poh HBr HF HClO CH2O NaOH NH OH Answer: If one of the four values ([H +, [OH -, ph or poh) is known, then the other three values can be calculated. The value of w at the given temperature must be known. In these examples, the temperature is 25 C and the value of w is Solution [H + (M) ph [OH - (M) poh HBr HF HClO CH2O NaOH NH OH

7 To convert ph to [H + [H + = 10 -ph To convert [H + to ph ph = -log[h + To convert ph to poh To convert poh to ph To convert poh to [OH - poh = 14 - ph ph = 14 - poh [OH = 10 -poh To convert [OH - to poh poh = -log[oh - To convert between [H + and [OH - w = [H + [OH - at 25 C = [H + [OH - 1. Question: Determine the ph for each of the given solutions. a M OH b M Ba(OH)2 c M NH3, b = Answer: To find the ph of these basic solutions, the poh needs to be calculated first. The approach will be different for strong and weak bases. a. OH is a strong base. For M OH, the [OH - = [OH therefore [OH - = M poh log[oh poh log[0.250 poh With the value of the poh, the ph can be determined by subtracting from 14. ph poh 14 ph ph 14 poh ph Given that OH is a strong base, a high ph value (close to 14) is expected.

8 b. For M Ba(OH)2, the approach will be similar except for the fact that there are two hydroxide groups for every unit of Ba(OH)2 so the hydroxide concentration is twice that of the base. Therefore [OH - = M poh log[oh poh log[0.500 poh With the value of the poh, the ph can be determined by subtracting from 14. ph poh 14 ph ph 14 poh ph Given that the hydroxide concentration is higher for Ba(OH)2 than for OH, the ph will be higher (more basic). c M NH3, b = Ammonia is a weak base. Therefore, an equilibrium equation and ICE table must be used to determine the poh, and then the ph of the solution. The b value (givn in the problem) is NH3(aq) + H2O(l) NH4 + (aq) + OH - (aq) INITIAL: The initial concentration of NH3 is given as M. The initial concentrations of NH4 + and OH - are zero and the concentration of water is omitted. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column.

9 NH3 (M) H2O (M) NH4 + (M) OH - (M) I M 0 0 C -x +x +x E x x x The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [OH b -5 - NH OH 4 NH3 x x x The value of b is small, so the assumption that x << can be made. -5 x x Now, the value of x can be determined. x x The value of x is less than 5% of the value from which it was subtracted (0.250) so the assumption is valid. The equilibrium value of [OH - = x, therefore [OH - = The poh of the solution can now be calculated and from that, the ph of the solution since ph + poh = 14 (at 25 C). poh log[oh poh poh log( ) ph poh ph ph Given that NH3 is a weak base, a ph value greater than 7 is expected so the answer is reasonable.

10 Ions as Acids and Bases 1. Question: Label each of the following ions as acidic, basic, or neutral. a. Cl - d. SO4 2- b. NO2 - e. + c. NH4 + f. NO3 - Answer: a. Cl - : The anion is neutral because it does not react with water to any significant amount. Neutral anions can be recognized because there parent acids are strong acids. If Cl - reacted with water, the only product could be HCl which completely ionizes in water, therefore no H + or OH - ions are generated. b. NO2 - : The anion is basic and it will react with water to produce hydroxide ions. Anions whose parent acid is a weak acid will be basic because they can react with water to produce the acid (which as a weak acid does not completely ionize). The formation of some of the weak acid results in the formation of hydroxide ions. NO2 - (aq) + H2O(l) HNO2(aq) + OH - (aq) c. NH4 + : The cation is an acid and reacts with water to produce hydronium ions. The reaction with water will also produce NH3, which is a weak base. NH4 + (aq) + H2O(l) NH3(aq) + H3O + (aq) d. SO4 2- : The anion is basic because it will react with water to produce HSO4 2-. The parent acid is HSO4 - which is a weak acid. The formation of the weak acid upon reaction of SO4 2- and H2O also results in the formation of hydroxide ions. SO4 2- (aq) + H2O(l) HSO4 - (aq) + OH - (aq) e. + : The cation is neutral because it will not react with water to produce either H + or OH - ions. f. NO3 - : The anion is neutral because will not react with water to produce either H + or OH - ions. The parent acid of NO3 - is HNO3 which is a strong acid. If the nitrate ion reacted with water, the possible product is a strong acid which completely dissociates in water.

11 2. Question: Determine whether solutions of the following salts are acidic, basic, or neutral. a. NaCl b. NO3 c. NO2 d. NH4Cl e. NH4F Answer: a. NaCl: Neutral; Na + is a neutral cation and Cl - is a neutral anion so the resulting salt is neutral. b. NO3: Neutral; + is a neutral cation and NO3 - is a neutral anion so the resulting salt is neutral. c. NO2: Basic; + is a neutral cation and NO2 - is a basic anion so the resulting salt is basic. d. NH4Cl: Acidic; NH4 + is an acidic cation and Cl - is a neutral anion so the resulting salt is acidic. e. NH4F: Acidic; NH4 + is an acidic cation and F - is a basic anion. The acidbase nature of the salt will be determined by the relative strengths of the acidic cation and basic anion. The a of HF is and the b of NH3 is HF is the stronger of the two so it will dominate and the resulting solution is acidic. 3. Question: Calculate the ph of a M solution of NO2. The a of HNO2 is Answer: The ph of a M NO2 solution can be calculated. Given that it is an ionic compound, two reactions have to be considered; the reaction of + or NO2 - with water. + (aq) + H2O(l) No Reaction If + reacts with water, the only possible product is OH, a strong base that completely dissociates. Therefore, the presence of + will not affect the ph of the solution. NO2 - (aq) + H2O(l) HNO2(aq) + OH - (aq)

12 The nitrite ion (NO2 - ) reacts with water to produce HNO2, a weak acid. Since HNO2 is a weak acid, it does not completely dissociate upon formation which will result in an increase in the [OH - and therefore change the ph of the solution. This is the only reaction that needs to be considered to determine the ph of the NO2 solution. In the reaction, NO2 - behaves as a base (proton acceptor) but the the a value of HNO2 ( ) can be obtained from tables. Therefore, w = a b must be used to determine the value of b w b a b b Now that the reaction and the equilibrium constant are known, an ICE table can be constructed to find the ph of the solution. INITIAL: The initial concentration of NO2 - is given as M. The initial concentrations of HNO2 and OH - are zero and the concentration of water is omitted. CHANGE: The change is represented by x and follows the stoichiometry of the balanced chemical reaction. EQUILIBRIUM: The equilibrium row is the sum of the initial and change rows for each column. NO2 - (M) H2O (M) HNO2 (M) OH - (M) I M 0 0 C -x +x +x E x x x The law of mass action is given and the equilibrium values may be substituted into the equation to solve for x and thus the value of [OH.

13 2.310 b HNO2 OH NO 2 x x x The value of b is small, so the assumption that x << can be made. -11 x x Now, the value of x can be determined. x 2-6 x The value of x is less than 5% of the value from which it was subtracted (0.500) so the assumption is valid. The equilibrium value of [OH - = x, therefore [OH - = The poh of the solution can now be calculated and from that, the ph of the solution since ph + poh = 14 (at 25 C) poh log[oh poh poh log( ) ph poh ph ph 8.53 Given that NO2 - is acting as a weak base, a ph value greater than 7 is expected so the answer is reasonable.