Acids, Bases, and ph. ACIDS, BASES, & ph

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1 I. Arrhenius Acids and Bases ACIDS, BASES, & ph Acid any substance which delivers hydrogen ion (H + ) _ to the solution. Base any substance which delivers hydroxide ion (OH ) to the solution. II ph ph is defined as the negative logarithm of the hydrogen or hydronium ion concentration. ph = log [H + ] = log [H 3 O + ] Exercise 1: For each hydrogen ion concentration listed, find the ph. (List answers to the proper number of significant figures.) Also identify each as an acidic or a basic solution. 1) 3.27 x 10 4 M ph = log 3.27 x 10 4 = acidic. 2) 5.0 x 10 9 M ph = log 5.0 x 10 9 = 8.30 basic The ion product for water, Kw = [H 3 O + ][OH ] = 1.00 x Exercise 2: For each of the solutions in exercise 1, determine the concentration of hydroxide ion. 1) [OH ] = 1.00 x /3.27 x 10 4 [OH ] = M 2) [OH ] = 1.00 x /5.0 x 10 9 [OH ] = 2.0 x 10 6 M The poh of a solution is defined as poh = log[oh ] Exercise 3: For solutions in exercises 1 and 2 (they are the same ones), calculate the value of poh. For solution 1 practice using the relationship that ph + poh = For solution 2 practice using the relationship that poh = log[oh ]. SHOW YOUR CALCULATIONS. 1) poh = = ) poh = log 2.0 x 10 6 poh = 5.70 To convert back to the concentration of hydronium ion from the ph value, [H 3 O + ] = 10 ph Exercise 4: For solutions with the following ph values, determine the concentration of hydrogen ion (or hydronium ion) in solution. 1) ph = [H + ] = [H + ] = 3.4 x M 2) ph = 8.23 [H + ] = [H + ] = 5.9 x 10 9 M D61

2 III. Conjugate Acid Base Pairs The formulas for a conjugate acid/base pair are different by only one hydrogen ion. The conjugate base of a given acid has the same formula minus a hydrogen ion. The conjugate acid of a given base will have the same formula plus one hydrogen ion. Exercise 5. List the conjugates for the following acids and bases: Acid Conj. Base Base Conj. Acid CH 3 CO 2 H CH 3 CO 2 HSO 4 H 2 SO 4 HCN CN C 2 H 3 O 2 HC 2 H 3 O 2 H 2 O OH H 2 O H 3 O + H 2 PO 4 HPO 4 2 SO 4 2 HSO 4 IV. Relative Strengths of Acids and Bases The stronger the acid the weaker is it s conjugate base. The weaker the acid the stronger is it s conjugate base. The stronger the base the weaker is it s conjugate acid. The weaker the base the stronger is it s conjugate acid. Ka Values Table HSO4 SO x 10 2 HSO3 SO x 10 8 H2PO4 HPO x 10 8 HCO3 CO x HPO4 2 PO x HS S x Exercise 6. The strength of an acid is based on it s Ka value. The larger the value of Ka, the more it forms the product ion. This means that it is a stronger electrolyte and therefore a stronger acid. a) Given the Ka values above, find the strongest acid. (Ans: b) a) HCO3 c) H2PO4 e) HS b) HSO4 d) HSO3 b) Given the Ka values above, find the strongest base. (Ans: e) a) SO3 2 c) CO3 2 e) S 2 b) HPO4 2 d) SO4 2 c) HS is a acid than HPO4 2 and S 2 is a base than PO4 3. (Ans: a) a) weaker, stronger c) stronger, stronger b) stronger, weaker d) weaker, weaker d) Write a balanced proton transfer reaction between carbonate ion and hydrogen sulfate ion. Also identify the conjugate acid/base pairs. CO3 2 (aq) + HSO4 (aq) HCO3 (aq) + SO4 2 (aq) base acid conj. Acid conj. Base D62

3 e) Is the previous reaction product favored or reactant favored at equilibrium? (Remember that the equilibrium lies to the side of the weaker acid and base!) Which was the weaker base? SO4 2 Which was the weaker acid? HCO3 Is the reaction reactant or product favored? product V. Acid Strength Exercise 7 Write a Ka (hydrolysis) reaction for the following acids. Use the appropriate type of arrow, based on the strength of the acid. Strong acids go to completion while weak acids form an equilibrium.) HBrO 3 : HClO 4 : HBr: HNO 2 : HBrO 3(aq) + H 2 O (l) H 3 O + (aq) + BrO 3 (aq) HClO 4(aq) + H 2 O (l) H 3 O + (aq) + ClO 4 (aq) HBr (aq) + H 2 O (l) H 3 O + (aq) + Br (aq) HNO 2(aq) + H 2 O (l) H 3 O + (aq) + NO 2 (aq) Exercise 8 Calculate the Ka and % dissociation (aka % ionization) for the following: HCNO [H 3 O + ]equilib = 4.55 x 10 3 initial concentration of the acid = M HCNO + H 2 O H 3 O + + CNO I C x +x +x E 0.100x x x x = 4.55 x 10 3 = ` = = Ka = ( ) 2 = x x 10 4 ( ) % Dissociation = ( /0.100) x 100 = 4.55% VI. Calculation of H3O+ concentration and ph in acids from Ka values The concentration of H 3 O + for weak acids requires writing a Ka reaction, setting up an ICE table, and substituting the equilibrium row of the ICE table into the Ka expression. The ph can then be calculated from the H 3 O + concentration using ph = log [H 3 O + ]. D63

4 Exercise 9. Write Ka (hydrolysis) reactions for the following acids. Calculate the H 3 O + concentration and ph for each. Remember that the process is very different for strong acids than weak acids. a) M HCl HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl (aq) [H 3 O + ] = 0.015M ph = log(0.015) = 1.82 b) M HCN (Ka = 4.9 x 10 9 ) HCN (aq) + H 2 O (l) H 3 O + (aq) + CN (aq) I C x +x +x E 0.015x x x Ka = 4.9 x 10 9 = x 2 x x [H 3 O + ] = SQRT [(4.9 x 10 9 ) (0.015)] [H 3 O + ] = SQRT (7.36 x ) = 8.57 x x 10 6 M Check Approximation [(8.57 x 10 6 ) / (0.015)] x 100 = 0.057% (error < 5%; approximation OK!) ph = log(8.57 x 10 6 ) = 5.07 VII. Base Strength Strong Bases completely react with water to produce hydroxide ions while weak bases do not. Strong bases include Group 1 hydroxides (LiOH, NaOH, KOH, etc.) and Heavy Group 2 hydroxides {Ca(OH) 2, Ba(OH) 2, etc} Strong bases are like strong acids in that the final hydroxide concentration is equal to the initial base concentration (times the number of hydroxide ions in the formula). D64

5 VIII. Base Equilibria Just as with acids, strong bases react with water to produce 100% products while weak bases do not. For weak bases, the reaction always produces hydroxide ions and the conjugate acid of the base. Exercise 10 Write Kb (hydrolysis) reactions for the following bases. Use the proper type of arrow in your reaction. C 2 H 3 O 2 (aq) + H 2 O (l) HC 2 H 3 O 2(aq) + OH (aq) H 2 O LiOH (s) + H 2 O (l) Li + (aq) + OH (aq) + H 2 O (l) becomes LiOH (s) Li+ (aq) + OH (aq) F (aq) + H 2 O (l) HF (aq) + OH (aq) Exercise 11 Calculate the ph of the following solutions of strong bases. a) 3.7 x 10 4 M KOH KOH(aq) K + (aq) + OH (aq) [OH ] = [KOH] = 3.7 x 10 4 M poh = log(3.7 x 10 4 M) = 3.43 ph = = b) M Ba(OH) 2 Ba(OH) 2 (aq) Ba 2+ (aq) + 2 OH (aq) [OH ] = 2[Ba(OH) 2 ] = M (based on reaction stoichiometry) poh = log(0.0050) = 2.30 ph = = IX. Calculation of OH concentration in bases from Kb values The concentration of hydroxide ion in a solution of a weak base can be calculated using Kb in the same way that the hydronium ion concentration can be calculated in a weak acid using Ka. Unfortunately, tables of Kb values are usually much less extensive that those of Ka values. However, the value of Kb for a weak base can be calculated if the Ka of the conjugate acid is known using the formula: Kw = 1.00 x = (Ka)(Kb) D65

6 Exercise 12 For a M solution of sodium carbonate: a) Write the Kb reaction: CO 2 3 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) b) Determine the value of Kb (H2CO3: Ka1= 4.3 x 10 7, Ka2= 5.6 x ) K b = 1.0 x = 1.0 x = x x 10 4 Ka2 5.6 x c) Calculate the [OH] by c1) set up the ICE table CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) I C x +x +x E x x x c2) solve using the simplification approximation K b = x 10 4 = x 2 x 2 (based on approximation simplification) ( x) ( ) x 2 = ( )(1.786 x 10 4 ) = x 10 7 X = SQRT(6.161 x 10 7 ) = x 10 4 M = [OH ] c3) Check approximation: 7.9 x 10 4 M [(7.849 x 10 4 ) / ( )] x 100 = 22.8% (error > 5%; approximation failed) c4) Solve using the quadratic A = 1 (1.786 x 10 4 )( x)=x 2 B = x 10 4 C = x 10 7 (6.162 x 10 7 )+(1.786 x 10 4 x) = x 2 X = or x x 10 4 x x 10 7 = 0 [OH ] = M or 7.0 x 10 4 M d) Determine the ph of the solution. poh = =log( ) = 3.15 ph = D66