Tang 04 ka calculations 2

K a  Calculations
K a  Calculations Ionization Constant (K a ) K a  – ionization constant for a weak acid HC 2 H 3 O 2(aq)  <===> H + (aq)  + C 2 H 3 O 2 - (aq) What is the formula for K a ? K a  = [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ]
K a  Calculations Ionization Constant (K a ) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
K a  Calculations pK a How are K a  and pK a  related? pK a  = - log K a small pK a  = more ionization; stronger acid large pK a  = less ionization; weaker acid
K a  Calculations Example #1 ,[object Object],[object Object],pK a  = -log K a pK a  = -log (1.8x10 -5 ) pK a  = 4.7 pK a  = -log K a K a  = 10 -pK a K a  = 10 –(9.24) K a  = 5.75x10 –10
K a  Calculations Example #2 Hypoiodous acid has a pK a  of 10.6.  The pK a  of hypobromous acid is 8.64.  What is the chemical formula for each substance?  Which is the weaker acid? Hypoiodous acid =  HIO (aq) pK a  = 10.6 Hypobromous acid =  HBrO (aq) pK a  = 8.64 K a  = 10 -pK a K a  = 10 –(10.6) K a  = 2.51x10 –11 K a  = 10 -pK a K a  = 10 –(8.64) K a  = 2.00x10 –9 Smaller value Smaller pK a  = stronger acid Check out the k a …
K a  Calculations Calculations using pH ,[object Object],[object Object],[object Object]
K a  Calculations Example #3 Formic acid, HCHO 2  is a monoprotic acid.  In a 0.100 M solution of formic acid, the pH is 2.38 at 25°C.  Calculate the K a  and pK a  for formic acid at this temperature.
K a  Calculations EXAMPLE #3: Formic acid, HCHO 2  is a monoprotic acid.  In a 0.100 M solution of formic acid, the pH is 2.38 at 25°C.  Calculate the K a  and pK a  for formic acid at this temperature. [H + ] = 10 -pH [H + ] = 10 -(2.38) [H + ] = 0.004168694M HCHO 2(aq)  <===> H + (aq)  + CHO 2 - (aq)    at equilibrium, x = [H + ] = [CHO 2 ] = 0.004168694M I  0.100M 0 0 C  -x +x +x E  0.100-x x x 0.0958   0.004169 0.004169    0.100-x = 0.100 – (0.004168694) = 0.0958343 K a  =  [H + (aq) ][CHO 2 - (aq) ] [HCHO 2(aq) ]  K a  =  [0.004169][0.004169] [0.0958]  K a  = 0.000181425  pK a  = -log K a pK a  = -log (0.000181425) pK a  = 3.74 .: k a  = 1.81x10 -4  and pK a  = 3.74
K a  Calculations Example #4 A 0.100 M solution of the weak acid HF was found to have an [H 3 O+] = 0.0080 M at equilibrium.  Calculate the K a  and pK a  for HF.
K a  Calculations Example #4: A 0.100 M solution of the weak acid HF was found to have an [H 3 O+] = 0.0080 M at equilibrium.  Calculate the K a  and pK a  for HF. I  0.100M 0 0 C  -x +x +x E  0.100-x x x HF (aq)  <===> H + (aq)  + F - (aq) 0.008 0.008 0.092 K a  =  [H + (aq) ][F - (aq) ] [HF (aq) ]  K a  =  [0.008][0.008] [0.092]  K a  = 0.000695652  pK a  = -log K a pK a  = 3.1576 .: HF has a k a  of 7.0x10 -4  and a pK a  of 3.2
K a  Calculations Example #5 A student planned an experiment that would use 0.100 M acetic acid.  Calculate the values of [H + ] and pH.  K a  = 1.8 x 10 -5
K a  Calculations Example #5: A student planned an experiment that would use 0.100 M acetic acid.  Calculate the values of [H + ] and pH.  K a  = 1.8 x 10 -5 HC 2 H 3 O 2(aq)  <===> H + (aq)  + C 2 H 3 O 2 - (aq) I  0.100M   0 0 C  -x   +x +x E  0.100-x   x x K a  =  [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ]  1.8x10 -5  =  [x][x] [0.100-x]  1.8x10 -5  (0.100) =  x 2 1.8x10 -6  =  x 2 Assumption    0.100-x = 0.100 0.00134 =  x
K a  Calculations Example #5 continued: A student planned an experiment that would use 0.100 M acetic acid.  Calculate the values of [H + ] and pH.  K a  = 1.8 x 10 -5 HC 2 H 3 O 2(aq)  <===> H + (aq)  + C 2 H 3 O 2 - (aq) I  0.100M   0 0 C  -x   +x +x E  0.100-x   x x x = 0.00134 = [H + ] pH = -log [H + ] pH = -log [0.00134] pH = 2.87 .: [H + ] = 1.34x10 -3 M and pH = 2.87
K a  Calculations Example #6 What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution?  pKa = 4.74
K a  Calculations Example #6: What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution?  pKa = 4.74 HC 2 H 3 O 2(aq)  <===> H + (aq)  + C 2 H 3 O 2 - (aq) I  0.25M   0 0 C  -x   +x +x E  0.25-x   x x K a  = 10 -pK a K a  = 10 –(4.74) K a  = 0.000018197 K a  =  [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ]  1.8197x10 -5  =  [x][x] [0.25-x]  Assumption    0.25-x = 0.25 1.8197x10 -5  =  [x][x] [0.25]  1.8197x10 -5  (0.25) = x 2 0.000004549 = x 2 0.00213 = x
K a  Calculations Example #6: What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution?  pKa = 4.74 HC 2 H 3 O 2(aq)  <===> H + (aq)  + C 2 H 3 O 2 - (aq) I  0.25M   0 0 C  -x   +x +x E  0.25-x   x x x = [H + ] = 0.00213 pH = -log [H + ] pH = -log [0.00213] pH = 2.67 .: the pH is 2.67
K a  Calculations Polyprotic Acids ,[object Object],[object Object],[object Object],[object Object]
K a  Calculations Each H +  released is associated with a K a  value. H 2 SO 4     H +  + HSO 4 - K a1  = very large HSO 4 -  <===> H +  + SO 4 2- K a2  = 1.0 x 10 -2 Why is H 2 SO 4  considered a strong acid? Because it’s first proton dissociates readily in water (has a very large k a  value)
K a  Calculations Example #7 Calculate the pH and [C 6 H 6 O 6 ] 2-  of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1  = 7.9 x 10 -5 K a2  = 1.6 x 10 -12
K a  Calculations Example #7: Calculate the pH and [C 6 H 6 O 6 ] 2-  of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1  = 7.9 x 10 -5 K a2  = 1.6 x 10 -12 H 2 C 6 H 6 O 6(aq)  <===> H + (aq)  + HC 6 H 6 O 6 - (aq) I  0.10M   0 0 C  -x   +x +x E  0.10-x   x x K a  =  [H + (aq) ][HC 6 H 6 O 6 - (aq) ] [H 2 C 6 H 6 O 6(aq) ]  7.9x10 -5  =  [x][x] [0.10]    assumption used 7.9x10 -6  =  x 2 0.002810694 =  x 0.00281 0.00281 0.09719 These will be used for the second H +  that is released
K a  Calculations Example #7 continued: Calculate the pH and [C 6 H 6 O 6 ] 2-  of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1  = 7.9 x 10 -5 K a2  = 1.6 x 10 -12 I  0.00281M  0.00281  0 C  -x   +x  +x E  0.00281-x 0.00281+x  x K a  =  [H + (aq) ][C 6 H 6 O 6 2 - (aq) ] [HC 6 H 6 O 6 - (aq) ]  1.6x10 -12  =  [0.00281+x][x] [0.00281-x] 1.6x10 -12  (0.00281-x)= 0.00281x+x 2 4.496x10 -15 -1.6x10 -12 x= 0.00281x+x 2 0 = x 2  +   0.00281x – 4.496x10 -15 x = 1.6x10 -12  or -0.00281   HC 6 H 6 O 6 - (aq)  <===> H + (aq)  + C 6 H 6 O 6 2- (aq)
K a  Calculations Example #7 continued: Calculate the pH and [C 6 H 6 O 6 ] 2-  of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1  = 7.9 x 10 -5 K a2  = 1.6 x 10 -12 I  0.00281M   0.00281 0 C  -x   +x +x E  0.00281-x   0.00281+x x HC 6 H 6 O 6 - (aq)  <===> H + (aq)  + C 6 H 6 O 6 2- (aq) 0.00281 0.00281 x = 1.6x10 -12 1.6x10 -12 pH = -log [H + ] pH = -log [0.00281] pH = 2.55 .: pH = 2.6 and [C 6 H 6 O 6 2- (aq) ] = 1.6x10 -12 M
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Tang 04 ka calculations 2

  • 1. K a Calculations
  • 2. K a Calculations Ionization Constant (K a ) K a – ionization constant for a weak acid HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) What is the formula for K a ? K a = [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ]
  • 3.
  • 4. K a Calculations pK a How are K a and pK a related? pK a = - log K a small pK a = more ionization; stronger acid large pK a = less ionization; weaker acid
  • 5.
  • 6. K a Calculations Example #2 Hypoiodous acid has a pK a of 10.6. The pK a of hypobromous acid is 8.64. What is the chemical formula for each substance? Which is the weaker acid? Hypoiodous acid = HIO (aq) pK a = 10.6 Hypobromous acid = HBrO (aq) pK a = 8.64 K a = 10 -pK a K a = 10 –(10.6) K a = 2.51x10 –11 K a = 10 -pK a K a = 10 –(8.64) K a = 2.00x10 –9 Smaller value Smaller pK a = stronger acid Check out the k a …
  • 7.
  • 8. K a Calculations Example #3 Formic acid, HCHO 2 is a monoprotic acid. In a 0.100 M solution of formic acid, the pH is 2.38 at 25°C. Calculate the K a and pK a for formic acid at this temperature.
  • 9. K a Calculations EXAMPLE #3: Formic acid, HCHO 2 is a monoprotic acid. In a 0.100 M solution of formic acid, the pH is 2.38 at 25°C. Calculate the K a and pK a for formic acid at this temperature. [H + ] = 10 -pH [H + ] = 10 -(2.38) [H + ] = 0.004168694M HCHO 2(aq) <===> H + (aq) + CHO 2 - (aq)  at equilibrium, x = [H + ] = [CHO 2 ] = 0.004168694M I 0.100M 0 0 C -x +x +x E 0.100-x x x 0.0958 0.004169 0.004169  0.100-x = 0.100 – (0.004168694) = 0.0958343 K a = [H + (aq) ][CHO 2 - (aq) ] [HCHO 2(aq) ] K a = [0.004169][0.004169] [0.0958] K a = 0.000181425 pK a = -log K a pK a = -log (0.000181425) pK a = 3.74 .: k a = 1.81x10 -4 and pK a = 3.74
  • 10. K a Calculations Example #4 A 0.100 M solution of the weak acid HF was found to have an [H 3 O+] = 0.0080 M at equilibrium. Calculate the K a and pK a for HF.
  • 11. K a Calculations Example #4: A 0.100 M solution of the weak acid HF was found to have an [H 3 O+] = 0.0080 M at equilibrium. Calculate the K a and pK a for HF. I 0.100M 0 0 C -x +x +x E 0.100-x x x HF (aq) <===> H + (aq) + F - (aq) 0.008 0.008 0.092 K a = [H + (aq) ][F - (aq) ] [HF (aq) ] K a = [0.008][0.008] [0.092] K a = 0.000695652 pK a = -log K a pK a = 3.1576 .: HF has a k a of 7.0x10 -4 and a pK a of 3.2
  • 12. K a Calculations Example #5 A student planned an experiment that would use 0.100 M acetic acid. Calculate the values of [H + ] and pH. K a = 1.8 x 10 -5
  • 13. K a Calculations Example #5: A student planned an experiment that would use 0.100 M acetic acid. Calculate the values of [H + ] and pH. K a = 1.8 x 10 -5 HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) I 0.100M 0 0 C -x +x +x E 0.100-x x x K a = [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ] 1.8x10 -5 = [x][x] [0.100-x] 1.8x10 -5 (0.100) = x 2 1.8x10 -6 = x 2 Assumption  0.100-x = 0.100 0.00134 = x
  • 14. K a Calculations Example #5 continued: A student planned an experiment that would use 0.100 M acetic acid. Calculate the values of [H + ] and pH. K a = 1.8 x 10 -5 HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) I 0.100M 0 0 C -x +x +x E 0.100-x x x x = 0.00134 = [H + ] pH = -log [H + ] pH = -log [0.00134] pH = 2.87 .: [H + ] = 1.34x10 -3 M and pH = 2.87
  • 15. K a Calculations Example #6 What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution? pKa = 4.74
  • 16. K a Calculations Example #6: What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution? pKa = 4.74 HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) I 0.25M 0 0 C -x +x +x E 0.25-x x x K a = 10 -pK a K a = 10 –(4.74) K a = 0.000018197 K a = [H + (aq) ][C 2 H 3 O 2 - (aq) ] [HC 2 H 3 O 2(aq) ] 1.8197x10 -5 = [x][x] [0.25-x] Assumption  0.25-x = 0.25 1.8197x10 -5 = [x][x] [0.25] 1.8197x10 -5 (0.25) = x 2 0.000004549 = x 2 0.00213 = x
  • 17. K a Calculations Example #6: What pH results when 0.25 mol of acetic acid is dissolved in enough water to make 1.00 L of solution? pKa = 4.74 HC 2 H 3 O 2(aq) <===> H + (aq) + C 2 H 3 O 2 - (aq) I 0.25M 0 0 C -x +x +x E 0.25-x x x x = [H + ] = 0.00213 pH = -log [H + ] pH = -log [0.00213] pH = 2.67 .: the pH is 2.67
  • 18.
  • 19. K a Calculations Each H + released is associated with a K a value. H 2 SO 4  H + + HSO 4 - K a1 = very large HSO 4 - <===> H + + SO 4 2- K a2 = 1.0 x 10 -2 Why is H 2 SO 4 considered a strong acid? Because it’s first proton dissociates readily in water (has a very large k a value)
  • 20. K a Calculations Example #7 Calculate the pH and [C 6 H 6 O 6 ] 2- of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1 = 7.9 x 10 -5 K a2 = 1.6 x 10 -12
  • 21. K a Calculations Example #7: Calculate the pH and [C 6 H 6 O 6 ] 2- of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1 = 7.9 x 10 -5 K a2 = 1.6 x 10 -12 H 2 C 6 H 6 O 6(aq) <===> H + (aq) + HC 6 H 6 O 6 - (aq) I 0.10M 0 0 C -x +x +x E 0.10-x x x K a = [H + (aq) ][HC 6 H 6 O 6 - (aq) ] [H 2 C 6 H 6 O 6(aq) ] 7.9x10 -5 = [x][x] [0.10]  assumption used 7.9x10 -6 = x 2 0.002810694 = x 0.00281 0.00281 0.09719 These will be used for the second H + that is released
  • 22. K a Calculations Example #7 continued: Calculate the pH and [C 6 H 6 O 6 ] 2- of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1 = 7.9 x 10 -5 K a2 = 1.6 x 10 -12 I 0.00281M 0.00281 0 C -x +x +x E 0.00281-x 0.00281+x x K a = [H + (aq) ][C 6 H 6 O 6 2 - (aq) ] [HC 6 H 6 O 6 - (aq) ] 1.6x10 -12 = [0.00281+x][x] [0.00281-x] 1.6x10 -12 (0.00281-x)= 0.00281x+x 2 4.496x10 -15 -1.6x10 -12 x= 0.00281x+x 2 0 = x 2 + 0.00281x – 4.496x10 -15 x = 1.6x10 -12 or -0.00281 HC 6 H 6 O 6 - (aq) <===> H + (aq) + C 6 H 6 O 6 2- (aq)
  • 23. K a Calculations Example #7 continued: Calculate the pH and [C 6 H 6 O 6 ] 2- of a 0.10 M solution of H 2 C 6 H 6 O 6(aq) . K a1 = 7.9 x 10 -5 K a2 = 1.6 x 10 -12 I 0.00281M 0.00281 0 C -x +x +x E 0.00281-x 0.00281+x x HC 6 H 6 O 6 - (aq) <===> H + (aq) + C 6 H 6 O 6 2- (aq) 0.00281 0.00281 x = 1.6x10 -12 1.6x10 -12 pH = -log [H + ] pH = -log [0.00281] pH = 2.55 .: pH = 2.6 and [C 6 H 6 O 6 2- (aq) ] = 1.6x10 -12 M