Solutions for the problems about „Calculation of …

1 Solutions for the problems about Calculation of pH in the case of monoprotic acids and bases 1. What is the pH of a M acetic acid solution? Acetic acid is a weak acid with Ka = 10 5 and in this case cweak acid >>> Ka, that is the equation to use is: [H+] =acidweak ac K= ) (5 = M pH = -log[H+] = -log( ) = 2. What is the pH of a M ammonia solution? Ammonia is a weak base with Kb = 10 5 and in this case cweak base >>> Kb, that is the equation to use is: [OH ] =baseweak bc K= ) (5 = M pOH = -log[OH-] = -log( ) = pH = pH = = 3. What is the pH of a M sodium acetate solution? Sodium acetate is a weak base, a conjugate base of acetic acid, so: acetatebacidaceticaKK = KW that is acidaceticaWacetatebKKK == = 10-10In this case cweak base >>> Kb, that is the equation to use is: [OH ] =baseweak bc K= ) (10 = 10-6 M pOH = -log[OH-] = -log( 10-6) = pH = pH = = 4. What is the concentration (in g/dm3 units) of an ammonia solution which has a pH of Ammonia is a weak base with Kb = 10 5.

2 PH = that is pOH = pH = [OH-] = 10 pOH = 10 = 10-3 M ][OH][OHc][OHbaseweak b =K 33baseweak = 10-6 = 10-5 (cweak base 10-3) 10-6 = 10-5 cweak base 10-6 = 10-5 cweak base cweak base = M So, we have mol ammonia in 1 L solution, that is g ammonia (m = n = mol 17 g/mol) in 1 L solution. The concentration is g/L. 5. A monobasic organic acid has a pK of The pH of a saturated solution of this acid is Calculate the solubility of this organic acid in mol/dm3 units. 1pH = , so [H+] = 10 pH = 10 = 10-4 M pK = , so Ka = 10 pK = 10 = 10-5 ][H][Hc][Hacidweak a+++ =K = acid weakc 10-8 = 10-5 (cweak acid 10-4) 10-8 = 10-5 cweak acid 10-8 = 10-5 cweak acidcweak acid = M 6. What is the pH and the degree of dissociation in a a) M; in a b) M and in a c) M acetic acid solution, respectively? a) Acetic acid is a weak acid, so: ][H][Hc][Hacidweak a+++ =K However, if the cweak acid = M, we can use the simpler form of the formula: acid weaka2c][H =+K [H+] = [H+] = 10-3 M pH = -log[H+] = -log( 10-3) = 2In this case [H+] = [acetate ion], so =acidweak c][H+= = % b)Acetic acid is a weak acid, so: ][H][Hc][Hacidweak a+++ =K )][Hc(][Hacidweak a2++ =K 0)c(-])[H(][Hacidweak aa2= +++KK By solving the equation, 2)K4KK2aacid weakaac(-][H + =+= 10-4 M pH = -log[H+] = -log( 10-4) = In this case [H+] = [acetate ion], so =acidweak c][H+= = % c) Here again we do not have enough difference in the order of magnitude to ][H][Hc][Hacidweak a+++ =K )][Hc(][Hacidweak a2++ =K 0)c(-])[H(][Hacidweak aa2= +++KK By solving the equation, 2)K4KK2aacid weakaac(-][H + =+= 10-4 M pH = -log[H+] = -log( 10-4) = In this case [H+] = [acetate ion], so =acidweak c][H+= = % 7.

3 What is the pH in a M solution of a moderately weak acid if the Ka = 10 1 ? In the case of this moderately weak acid we do not have enough difference in the order of magnitude as = = . So the following formula should be used: ][H][Hc][Hacidweak a+++ =K )][Hc(][Hacidweak a2++ =K0)c(-])[H(][Hacidweak aa2= +++KK 0) (-])[ (][H112= + + + By solving the equation, 2)K4KK2aacid weakaac(-][H + =+= 10 3 M pH = -log[H+] = -log( 10 3) = 8. A windscreen washing liquid contains ammonia in g/dm3 concentration. What is the pH of this liquid? 1 L solution contains g of NH3, that is mol of NH3 (n = 17g 2= mol). So the concentration of ammonia is: cammonia = L 1mol M In this case, cweak base >>> Kb, so [OH ] =baseweak bc K= = M pOH = -log[OH ] = -log( ) = , pH = pOH = 9. cm3 of M ammonia solution is titrated with M HClO4. What is the added volume of titrant and the pH at 75% degree of titration?

4 (6 cm3, ) The molar amount of ammonia: n = c V = M L = mol At 100 % titration, nHClO4 = nNH3 = mol At 75 % titration, nHClO4 = nNH3 = mol HClO4. We know the concentration ( M) of the HClO4 solution, so, the volume can be calculated: V = cn= L = 6 mL At 75 % titration, we have a buffer, NH3 and NH4Cl together. 75 % of the original NH3 amount is converted to NH4Cl, and 25 % is remained as NH3. So, using the equation for buffers: [OH-] = Kbsaltbaseweak nn = = = 10-6 M 3pOH = -log[OH ] = -log( 10-6) = , pH = pOH = 10. The concentration of a monochloro acetic acid solution is M. What are the pH and the degree of dissociation in this solution? Ka = 10-3 Monochloro acetic acid is a weak acid and we do not have enough difference in the order of magnitude so the following formula should be used: ][H][Hc][Hacidweak a+++ =K The cweak acid = M and Ka = 10-3][H][ ][H3++ + = 0) (-])[ (][H332= + + + By solving the equation, 2) ( ) ( ][H-323-3 + = + [H+] = 10 4 M pH = log[H+] = log ( 10 4) pH = In this case [H+] = [monochloro acetate ion], so =acidweak c][H+= = % 11.

5 Calculate the pH when we add: a) 0 mL b) 9 mL c) 20 mL d) 25 mL NaOH solution whose concentration is M to a 10 mL sample of acetic acid. The concentration of the acetic acid is unknown (Ka= 10 5). First, calculate the concentration of the acetic acid if we know that 20 mL of NaOH is consumed up to the equivalence point. What kind of indicator would you use for this titration? Results: a) c(acetic acid) = mol/dm3 b) at the beginning of the titration: 0 ml of base was added (weak acid) pH = c) After the addition: 9 ml of base (buffer system) pH = d) After the addition: 20 ml of base (weak base) pH = e) After the addition: 25 ml of base (excess of strong base) pH = 412. 10 mL of acetic acid, whose concentration is unknown is titrated with potassium hidroxide. a) Calculate the concentration of the acetic acid if the volume of potassium hidroxide that is consumed up to the equivalence point is 16 mL.

6 The concentration of KOH is M. b) Calculate the pH after the addition of 10 mL of KOH. Results: a) c(acetic acid) = M b) It will be a buffer system: pH = 13. A 10 mL sample of acetic acid with M concentration is titrated with sodium hidroxide. The concentration of sodium hidroxide is M as well. Decide whether it is right or wrong to use an indicator whose pK= Support your opinion with calculations! a) At first, we assume that the indicator marked the equivalence point correctly. This would mean that the same amounts of acetic acid and sodium hidroxide are present in the solution. n(acetic acid) = 10 = 1 mmol n(acetic acid) = n(NaOH) = 1 mmol This amount of NaOH is present in 10 mL of solution. Then you should calculate the pH of this solution (which contains a weak base, sodium acetate) to check your assumption about the indicator. Ka = 10 5 csalt/weak base = 201= M Kb= awKK= = 10 10 [OH ] =baseweak bc K= 10 6 M pOH = -log[OH ] = -log( 10 6) = pH = = From this result you can see that at the equivalent point the pH is , so the indicator was wrong.

7 B) Now, let s see where the indicator marked the end-point. Around pH , we still have a buffer system of acetic acid and sodium acetate. n(acetic acid) = 1 mmol n(NaOH) = y = mmol [H+] = 10 = 10 4 M [H+] = Ka()saltbase weakacid weaknn= 10 4 = 10 5 = 1 - = 1 y = mL Only this amount of sodium hidroxide is given to the acid when the indicator signs! So the mistake that you make by using the incorrect indicator is [( )/10] 100= % 14. 20 mL of benzoic acid of unknown concentration is titrated with potassium hydroxide whose concentration is M. Calculate the pH at: a) 0 %; b) 25 %; c) 100 %; d) 150 % degree of titration. Up to the equivalence point mL of KOH solution is consumed. (Ka = 10 5) a) n(KOH) = = mmol n(KOH) = n(benzoic acid) = mmol c(benzoic acid) = 10 2 M Ka = 10 5 As we do not have the three orders of magnitude difference between the concentration and the acidic ionization constant, the more complicated formula has to be used for the calculation of the pH.

8 ][H][Hc][Hacidweak a+++ =K )][Hc(][Hacidweak a2++ =K 0)c(-])[H(][Hacidweak aa2= +++KK 2)cK(4KK-][Hacidweak a2aa + =+ [H+] = () + = 10 3 M pH = -log[H+] = -log( 10 3) = b) n(KOH) = = mmol n(benzoic acid)sum = mmol We get a basic buffer solution of: n(benzoic acid)rest = mmol n(potassium benzoate) = mmol [H+] = Ka()saltbase weakacid weaknn= 10 5 10 4 M 6c) n(potassium benzoate) = mmol c(potassium benzoate) = = 10 2 M The salt, potassium benzoate is a weak base, so first the basic ionization constant should be calculated. Kb= awKK= = 10 10 In this case cweak base >>> Kb, that is the equation to use is: [OH ] =baseweak bc K= ) (10 = 10 6 M pOH = -log[OH ] = -log( 10 6) = pH = = d) n(KOH) = = mmol n(benzoic acid)sum = mmol n(potassium benzoate) = mmol n(KOH)excess = - = mmol As we have a weak base and a strong base in the system, it is the strong base, namely the KOH solution in excess that determines the pH of the solution.

9 C(KOH sol.) = 10 2 M pOH = -log[OH ] = -log( 10 2) = pH = = 15. 100 mL of formic acid whose concentration is M is titrated by sodium hydroxide solution of M concentration. Calculate the pH: a) before the titration b) after the addition of 8 mL; c) mL of NaOH solution; d) at the equivalence point. (Ka = 10 4) a) c(formic acid) = M Ka = 10 4 As we do not have the three orders of magnitude difference between the concentration and the acidic ionization constant, the more complicated formula has to be used for the calculation of the pH. ][H][Hc][Hacidweak a+++ =K )][Hc(][Hacidweak a2++ =K 70)c(-])[H(][Hacidweak aa2= +++KK 2)K4KK2aacid weakaac(-][H + =+()() + = + ][H= 10 3 M pH = -log[H+] = -log( 10 3) = b) n(formic acid) = 100 = mmol n(NaOH) = 8 = mmol After the addition of mL of NaOH solution, we get a basic buffer solution of HCOONa/HCOOH.

10 N(formic acid) = mmol n(sodium formiate) = mmol [H+] = Ka()saltbase weakacid weaknn= 10 4 10 5 M pH = -log[H+] = -log( 10 5) = c) n(formic acid) = 100 = mmol n(NaOH) = = mmol After the addition of mL of NaOH solution, we get a basic buffer solution of HCOONa/HCOOH. n(formic acid) = mmol n(sodium formiate) = mmol [H+] = Ka()saltbase weakacid weaknn= 10 4 10 6 M pH = -log[H+] = -log( 10 6) = d) n(formic acid) = 100 = mmol n(NaOH) = 10= mmol n(sodium formiate) = mmol The salt, sodium formiate is a weak base, so first the basic ionization constant should be calculated. Kb= awKK= = 10 11 c(sodium formiate) = ()solutionV)iatesodiumform(n= 1101= M In this case cweak base >>> Kb, that is the equation to use is: 8[OH ] =baseweak bc K= ) (11 = 10 7 M pOH = -log[OH ] = -log( 10 7) = pH = = 9