17.E: Aspectos adicionales de los equilibrios acuosos (ejercicios)

Respuestas

a. (PH = -log ([H_3O ^ {+}]) = – log (1.218) = – 0.086 )

([HCl] = [H ^ {+}] )

(M_ {1} V_ {1} = M_ {2} V_ {2} rightarrow (6.09 , M) (25 , mL) = (x) (125 , mL) rightarrow x = 1.218 )

b. (pH + pOH = 14 pH de la derecha = 14-pOH = 14-0.80 = 13 )

(pOH = -log ([OH ^ {-}]) = – log (0.159) = 0.80 )

([NaOH] = [OH ^ {-}] )

(M_ {1} V_ {1} = M_ {2} V_ {2} rightarrow (2.55 , M) (5.0 , mL) = (x) (80 , mL) rightarrow x = 0.159 )

(pH + pOH = 14 pH derecho = 14-pOH = 14-1.321 = 12.7 )

(HCl: 50.0 , mL times frac {1 , L} {1,000 , mL} times frac {0.225 , mol} {1 , L} = 0.01125 , mol )

(NaOH: 100.0 , mL times frac {1 , L} {1,000 , mL} times frac {0.184 , mol} {1 , L} = 0.0184 , mol )

Tenga en cuenta que el número de moles para (NaOH ) es mayor que (HCl ), por lo tanto, calculamos (pOH ).

(0.0184-0.01125 = 0.00715 , mol )

( frac {0.00715 , mol} {. 15 , L} = 0.0477 )

(pOH = -log ([OH ^ {-}]) = – log (0.0477) = 1.321 )

3. (M_ {1} V_ {1} = M_ {2} V_ {2} rightarrow (0.50 , M) (x , L) = (0.86 , M) (0.25 , L ) rightarrow x , L = 4.3 times 10 ^ {- 2} , L )

4. Solución de video

a. (pH = pK_a + log ( frac {C_ {2} H_ {3} NaO_ {2}} {C_ {2} H_ {4} O_ {2}}) = 4.76 + log ( frac {0.0935} { 0.0105}) = 5.71 )

(HCl: 100 , mL times frac {1 , L} {1,000 , mL} times frac {0.0105 , mol} {1 , L} = 0.0105 , mol )

(C_2H_3NaO_2: 100 , mL times frac {1 , L} {1,000 , mL} times frac {0.115 , mol} {1 , L} = 0.0115 , mol )

(C_ {2} H_ {3} NaO_ {2} , (s) + HCl , (aq) rightarrow C_ {2} H_ {4} O_ {2} , (aq) + NaCl , (s) )

(C_ {2} H_ {3} NaO_ {2} )

(HCl )

(C_ {2} H_ {4} O_ {2} )

(NaCl )

0,0115

0,0105

–

-0,0105

+0,0105

–

0,0935

0,0105

–

b. (pH = pK_a + log ( frac {C_ {2} H_ {3} NaO_ {2}} {C_ {2} H_ {4} O_ {2}}) = 4.76 + log ( frac {0.01} { 0.005}) = 5.05 )

(HCl: 100 , mL times frac {1 , L} {1,000 , mL} times frac {0.0105 , mol} {1 , L} = 0.005 , mol )

(C_2H_3NaO_2: 100 , mL times frac {1 , L} {1,000 , mL} times frac {0.115 , mol} {1 , L} = 0.015 , mol )

(C_ {2} H_ {3} NaO_ {2} , (s) + HCl , (aq) rightarrow C_ {2} H_ {4} O_ {2} , (aq) + NaCl , (s) )

(C_ {2} H_ {3} NaO_ {2} )

(HCl )

(C_ {2} H_ {4} O_ {2} )

(NaCl )

0,015

0,005

-0,005

+0,005

0,01

0,005

c. (pH = 14-pOH = 14 – (- log ( frac {0.0118-0.0109} {0.2 , L})) = 11.7 )

(CH_ {3} COOH: 0.1 , L veces 0.109 , M = 0.0109 , mol )

(NaOH: 0.1 , L veces 0.118 , M = 0.0118 , mol )

(CH_ {3} COOH , (aq) + NaOH , (aq) rightarrow H_ {2} O , (l) + CH_ {3} COO ^ {-} Na , (aq) )

(CH_ {3} COOH )

(NaOH )

(H_ {2} O )

(CH_ {3} COONa )

0,0109

0,0118

–

-0,0109

–

+0.0109

(9 veces 10 ^ {- 4} )

–

0,0109

d. (pH = 4.76 + log ( frac {0.0055} {0.0943}) = 3.53 )

(CH_ {3} COOH: 0.1 , L veces 0.998 , M = 0.0998 , mol )

(NaOH: 0.05 , L veces 0.110 , M = 0.0055 , mol )

(CH_ {3} COOH , (aq) + NaOH , (aq) rightarrow H_ {2} O , (l) + CH_ {3} COONa , (aq) )

(CH_ {3} COOH )

(NaOH )

(H_ {2} O )

(CH_ {3} COONa )

0,0998

0,0055

–

-0,0055

–

+0,0055

0,0943

–

0,0055

5. Solución de video

a. (pH = -log ( frac {0.9728} {0.2 , L}) = 4.86 )

(HCl: 100 , mL times frac {1 , L} {1,000 , L} times0.983 , M = 0.983 , mol )

(NaF: 100 , mL times frac {1 , L} {1,000 , L} times 0.102 , M = 0.0102 , mol )

(HCl , (aq) + NaF , (s) rightarrow NaCl , (aq) + HF , (aq) )

(HCl )

(NaF )

(NaCl )

(HF )

0,983

0,0102

–

-0,0102

–

-0,0102

0,9728

–

-0,0102

b. (pH = pK_a + log ( frac {[NaF]} {[HF]}) = 3.14 + log ( frac {0.00515} {0.00575}) = 3.09 )

(HCl: 50 , mL veces frac {1 , L} {1,000 , L} veces 0.115 , M = 0.00575 , mol )

(NaF: 100 , mL times frac {1 , L} {1,000 , L} times 0.109 , M = 0.0109 , mol )

(HCl , (aq) + NaF , (s) rightarrow NaCl , (aq) + HF , (aq) )

(HCl )

(NaF )

(NaCl )

(HF )

0,00575

0,0109

–

-0,00575

–

+0,00575

0,00515

–

0,00575

c. (pH = 14-pOH = 14 – (- log (0.0498-0.0106)) = 12.6 )

(HF: 100 , mL times frac {1 , L} {1,000 , L} times 0.106 , M = 0.0106 , mol )

(NaOH: 50 , mL veces frac {1 , L} {1,000 , L} veces 0.996 , M = 0.0498 , mol )

(HF , (aq) + NaOH , (aq) rightarrow NaF , (aq) + H_ {2} O , (aq) )

(HCl (aq) + NaOH (aq) NaCl derecho , (aq) + H_2O )

d. (PH = pK_a + log ( frac {[C_2H_3NaO_2]} {[CH_ {3} COOH]}) = 4.76 + log ( frac {0.0107} {0.04935}) = 4.10 ) [19459006 ]

(C_2H_3NaO_2: 0.1 , L veces 0.107 , M = 0.0107 )

(CH_3COOH: 0.05 , mL veces 0.987 , M = 0.04935 )

(CH_ {3} COOH , (aq) + H_ {2} O rightarrow C_2H_3O ^ {-} + H_3O ^ {+} )

(CaCO_ {3} , (aq) +2 , HCl , (aq) rightarrow CaCl_ {2} , (aq) + H_2O , (l) + CO_ {2} , ( g) )

(CO_ {3} ^ {2 -} , (aq) + H ^ {+} , (aq) rightleftharpoons HCO_ {3} ^ {-} , (aq) )

(HCO_ {3} ^ {2 -} , (aq) + H ^ {+} , (aq) rightleftharpoons H_ {2} CO_ {3} , (aq) )

(K_ {a1} = 4.3 veces 10 ^ {- 7} )

(K_ {a2} = 5.0 veces 10 ^ {- 11} )

A 0 mL de adición de (HCl ): (pH = 14-pOH = 14 – (- log ( sqrt {0.0125 times frac {10 ^ {- 14}} {5.0 times 10 ^ {-11}}}) = 11.20 )

(CaCO_ {3}: 250 , mg times frac {1 , g} {1,000 , mg} times frac {1 , mol} {100.0869 , g} times frac {1} {200 , mL times frac {1 , L} {1,000 , mL}} = 0.0125 , M )

A 5 ml de adición de (HCl ): (pH = -log (5 times 10 ^ {- 11}) + log ( frac {0.002} {0.0005}) = 10.9 )

(CaCO_ {3}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.005 , L veces 0.1 , M = 0.0005 , mol )

(CO_ {3} ^ {2 -} )

(H ^ {+} )

(HCO_ {3} ^ {-} )

0,0025

0,0005

-0,0005

0,0005

0,002

0,0005

A 10 ml de adición de (HCl ): (pH = -log (5 times 10 ^ {- 11}) + log ( frac {0.001} {0.0015}) = 10.5 )

(CaCO_ {3}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.01 , L veces 0.1 , M = 0.001 , mol )

(CO_ {3} ^ {2 -} )

(H ^ {+} )

(HCO_ {3} ^ {-} )

0,0025

0,001

-0,001

0,001

0,0015

0,001

Con 15 ml de adición de (HCl ): (pH = -log (5 times 10 ^ {- 11}) + log ( frac {0.001} {0.0015}) = 10.1 )

(CaCO_ {3}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.015 , L veces 0.1 , M = 0.0015 , mol )

(CO_ {3} ^ {2 -} )

(H ^ {+} )

(HCO_ {3} ^ {-} )

0,0025

0,0015

-0,0015

0,0015

0,001

0,0015

Con 20 ml de adición de (HCl ): (pH = -log (5 times 10 ^ {- 11}) + log ( frac {5 times 10 ^ {- 4}} {0.002} ) = 9.7 )

(CaCO_ {3}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.02 , L veces 0.1 , M = 0.002 , mol )

(CO_ {3} ^ {2 -} )

(H ^ {+} )

(HCO_ {3} ^ {-} )

0,0025

0,002

-0,002

0,002

(5 veces 10 ^ {- 4} )

0,002

Con 25 ml de adición de (HCl ): primer punto de equivalencia: (pH = frac {-log (K_ {a1}) – log (K_ {a2})} {2} = frac {- log (4.3 times 10 ^ {- 7}) – log (5.0 times 10 ^ {- 11})} {2} = 8.33 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.025 , L veces 0.1 , M = 0.0025 , mol )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

-0,0025

0,0025

Con 30 ml de adición de (HCl ): (pH = -log (4.3 times 10 ^ {- 7}) + log ( frac {0.0025} {0.0005}) = 7.07 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.03 , L veces 0.1 , M = 0.003 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,003

-0,0025

0,0025

(5 veces 10 ^ {- 4} )

0,0025

Con 35 ml de adición de (HCl ): (pH = -log (4.3 times 10 ^ {- 7}) + log ( frac {0.0025} {0.001}) = 6.76 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.035 , L veces 0.1 , M = 0.0035 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,0035

-0,0025

0,0025

0,001

0,0025

A 40 mL de adición de (HCl ): (pH = -log (4.3 times 10 ^ {- 7}) + log ( frac {0.0025} {0.003}) = 6.59 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.040 , L veces 0.1 , M = 0.004 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,004

-0,0025

0,0025

0,0015

0,0025

A 45 ml de adición de (HCl ): (pH = -log (4.3 times 10 ^ {- 7}) + log ( frac {0.0025} {0.002}) = 6.46 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.045 , L veces 0.1 , M = 0.0045 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,0045

-0,0025

0,0025

0,002

0,0025

A 50 ml de adición de (HCl ): segundo punto de equivalencia: (pH = 3.86 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.05 , L veces 0.1 , M = 0.005 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,005

-0,0025

0,0025

A 55 ml de adición de (HCl ): (pH = -log ( frac {0.003} {0.055}) = 1.26 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.055 , L veces 0.1 , M = 0.0055 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,0055

-0,0025

0,0025

0,003

0,0025

A 60 ml de adición de (HCl ): (pH = -log ( frac {0.0035} {0.06}) = 1.23 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.06 , L veces 0.1 , M = 0.006 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,006

-0,0025

0,0025

0,0035

0,0025

A 65 ml de adición de (HCl ): (pH = -log ( frac {0.0040} {0.065}) = 1.21 )

(HCO_ {3} ^ {-}: 0.2 , L veces 0.0125 , M = 0.0025 , mol )

(HCl: 0.06 , L veces 0.1 , M = 0.006 )

(HCO_ {3} ^ {-} )

(H ^ {+} )

(H_ {2} CO_ {3} )

0,0025

0,0065

-0,0025

0,0025

0,0040

0,0025

7. Solución de video

(M_1V_1 = M_2V_2 rightarrow V_2 = frac {M_1V_1} {V_2} = frac {(0.2 , M) (0.5 , L)} {12.0 , M} = 8.33 times10 ^ {- 3} , L ) Por lo tanto, diluya 8.33 mL de HCl 12.0 M a 500.0 mL.

(M_1V_1 = M_2V_2 rightarrow V_2 = frac {M_1V_1} {V_2} = frac {(0.288 , M) (0.05 , L)} {0.2 , M} = 7.20 veces 10 ^ { -3} , L )

(M_1V_1 = M_2V_2 rightarrow V_2 = frac {M_1V_1} {V_2} = frac {(0.288 , M) (0.05 , L)} {0.187 , M} = 7.70 veces 10 ^ { -2} )

a. (M_ {1} V_ {1} = M_ {2} V_ {2} rightarrow M_2 = frac {M_1V_1} {V_ {2}} = frac {(0.582 , M) (0.060 , L) } {0.0719 , L} = 0.49 , M )

b. (M_ {1} V_ {1} = M_ {2} V_ {2} rightarrow V_ {2} = frac {M_ {1} V_ {1}} {M_ {2}} = frac {(0.582 , M) (0.050 , L)} {0.49 , M} = 59.4 veces 10 ^ {- 2} , L )

Volumen de base añadida (ml)	10,0	30,0	40,0	45,0	50,0	55.0	65,0	75,0
pH	0,73	1,22	1,76	7	12,2	12,5	13,0	13,1

(NaOH , (aq) + HCl , (aq) rightarrow NaCl , (s) + H_ {2} O , (l) )

A 0 ml de base añadida: (pH = -log (0.288) = 0.54 )

A 10 ml de base añadida: (pH = -log ( frac {0.0144-0.00321} {0.06}) = 0.73 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.01 , L veces 0.321 , M = 0.00321 )

A 30 ml de base añadida: (pH = -log ( frac {0.0144-0.00963} {0.08}) = 1.22 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.03 , L veces 0.321 , M = 0.00963 )

A 40.0 mL de base agregada: (pH = -log ( frac {0.0144-0.01284} {0.09}) = 1.76 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.04 , L veces 0.321 , M = 0.01284 )

A 45,0 ml de base añadida: (pH = 7 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.045 , L veces 0.321 , M = 0.0144 )

Todo (HCl ) se neutralizará en el punto de equivalencia ([H ^ {+}] = [OH ^ {-}] ) y el pH de la solución es 7.

A 50.0 mL de base agregada: (pH = 14-pOH = 14-log ( frac {0.01605-0.0144} {0.1}) = 12.2 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.050 , L veces 0.321 , M = 0.01605 )

Con 55 ml de base añadida: (pH = 14-pOH = 14-log ( frac {0.01605-0.0144} {0.1}) = 12.5 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.055 , L veces 0.321 , M = 0.017655 )

A 65 ml de base añadida: (pH = 14-pOH = 14-log ( frac {0.020865-0.0144} {0.07}) = 13.0 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.065 , L veces 0.321 , M = 0.020865 )

A 75 ml de base añadida: (pH = 14-pOH = 14-log ( frac {0.024075-0.0144} {0.08}) = 13.1 )

(HCl: 0.05 , L veces 0.288 , M = 0.0144 )

(NaOH: 0.075 , L veces 0.321 , M = 0.024075 )

10.

El punto de equivalencia está en (pH ) 7 y esto ocurre en 0.0193 L.

(V_ {2} = frac {(0.156 , M) (0.025 , L)} {0.202} = 0.0193 , L )

11.

pH en el punto de equivalencia: (14 – (- log sqrt {( frac {10 ^ {- 14}} {10 ^ {- 3.75}} times frac {0.01205} {0.05 + 0.1227}} )) = 8.30 )

(0.05 , L veces 0.241 , M = 0.01205 , mol )

(V_2 = frac {(0.05 , L) (0.241 , M)} {0.0982} = 0.1227 , L )

Volumen de base añadida (ml)	0	5	10	15	20	25
pH	2,18	2,38	2,70	2,89	3,04	3.16

A 0 mL de base agregada: (pH = -log ([H ^ {+}]) = – log (2.52 times 10 ^ {- 6}) = 2.18 )

([H ^ {+}] = sqrt {10 ^ {- 3.75} times 0.241 , M} = 2.52 times 10 ^ {- 6} )

Con 5 ml de base añadida: (pH = 3.75 + log ( frac {4.91 times 10 ^ {- 4}} {0.01205-4.91 times 10 ^ {- 4}}) = 2.38 ) [ 19459006]

(KOH: 0.05 , L por 0.0982 , M = 4.91 por 10 ^ {- 4} , mol )

(CH_ {2} O_ {2}: 0.05 , L veces 0.241 , M = 0.01205 , mol )

Con 10 ml de base añadida: (pH = 3.75 + log ( frac {9.82 times 10 ^ {- 4}} {0.01205-9.82 times 10 ^ {- 4}}) = 2.70 ) [ 19459006]

(KOH: 0.01 , L veces 0.0982 , M = 9.82 veces 10 ^ {- 4} , mol )

(CH_ {2} O_ {2}: 0.05 , L veces 0.241 , M = 0.01205 , mol )

Con 15 ml de base añadida: (pH = 3.75 + log ( frac {0.001473} {0.01205-0.001473} = 2.89 )

(KOH: 0.015 , L veces 0.0982 , M = 0.001473 )

(CH_ {2} O_ {2}: 0.05 , L veces 0.241 , M = 0.01205 )

A 20 ml de base añadida: (pH = 3.75 + log ( frac {0.001964} {0.01205-0.001964}) = 3.04 )

(KOH: 0.02 , L veces 0.0982 , M = 0.001964 )

(CH_ {2} O_ {2}: 0.05 , L veces 0.241 , M = 0.01205 )

Con 25 ml de base añadida: (pH = 3.75 + log ( frac {0.002455} {0.01205-0.002455}) = 3.16 )

(KOH: 0.025 , L veces 0.0982 , M = 0.002455 , mol )

(CH_ {2} O_ {2}: 0.05 , L veces 0.241 , M = 0.01205 )

12. (M_1V_1 = M_2V_2 rightarrow V_2 = frac {M_1V1} {M_2} = frac {(0.430 , M) (0.05 , L)} {0.150 , M} = 1.42 veces 10 ^ {- 1} , L )

Se necesitan 143 ml para desprotonar completamente el grupo ácido carboxílico.

a. Se necesitan 143 mililitros adicionales de KOH para desprotonar el grupo amonio.

b. En el primer punto de equivalencia: (pH = frac {(pKa_ {1} + pKa_ {2})} {2} = 5.95 )

c. Se necesitan 143 ml de titulante para obtener una solución en la que la glicina no tiene carga eléctrica. El punto isoeléctrico de la glicina 5.95.

13.

(pH = pK_a + log ( frac {[C_5H_5N]} {[C_5H_5NH ^ {+}]}) = (14-8.77) + log ( frac {0.02996} {0.0063434}) = 5.90 )

(C_5H_5N , (aq) + HCl , (aq) rightarrow C_5H_5NH ^ {+} , (aq) + Cl ^ {-} , (aq) )

(HCl: 32.2 , mL times frac {1 , L} {1,000 , mL} times 0.197 , M = 0.0063434 , mol )

(C_5H_5N: 150 , mL times frac {1 , L} {1,000 , mL} times 0.242 , M = 0.0363 , mol )

(C_5H_5N )

(HCl )

(C_5H_5NH ^ {+} )

(Cl ^ {-} )

0,0363

0,0063434

–

-0,0063434

+0,0063434

–

0,02996

0,0063434

–

14. (pH = frac {(4.21 + 5.64)} {2} = 4.93 )

15. (pH = frac {(2.85 + 5.70)} {2} = 4.25 )

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