Tang 04 ka 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) ]

K a Calculations Ionization Constant (K a ) How do you suppose the K a values of strong acids compare with weak acids? Strong K a values are very large all available H + ions are released into solution Weak K a values are small (< 1.00) some H + ions are released into solution

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 Calculate pK a for acetic acid given K a = 1.8 x 10 -5 Calculate K a for ammonium ion given pKa = 9.24. 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 Two types of calculations: Calculate K a and pK a from the pH of its solution given initial concentration. Calculate pH or [H + ] of a solution given the initial concentration and K a or pK a .

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 acids which may release multiple H + ions into solution (i.e. H 2 SO 4 ) Polyprotic acids release H+ ions into solution one step at a time. H 2 SO 4  H + + HSO 4 - HSO 4 - <===> H + + SO 4 2-

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

Tang 04 ka calculations 2

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Tang 04 ka calculations 2