Cálculos teóricos de la valoración de acido debil mas base fuerte

1. Cálculos teóricos de la valoración de
Ácido débil más Base fuerte
CH3
COOH + NaOH = CH3
COONA +H2
O
NaOH= 0.0982N
𝑁1𝑉1 = 𝑁2𝑉2 𝑉1 =
(0.10038)(10)
0.0982
= 10.22𝑚𝑙
VOL=10ml
CH3COOH= 0.10038N
10.22ml punto de equivalencia teórico.
A) 0ml de NaOH ka=1.75x10-5
𝑃𝑘𝑎 = −𝑙𝑜𝑔1.75𝑋10 − 5 = 4.7569
En el matraz solo hay ácido acético
[𝐻] = √𝑘𝑎[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
[𝐻] = √(1.75x10 − 5)(0.10038) = 1.3253𝑥10 − 3
𝑝𝐻 = −𝑙𝑜𝑔[𝐻]
𝑝𝐻 = − log(1.3253𝑥10 − 3) = 2.9 pH inicial
B) 1ml de NaOH
CH3COOH + NaOH = CH3COONa H2O
Meqinicial 1.0038 0.0982 ------- ------
Meqfinal 0.9056 ------ 0.0982 ------
VOLUMEN TOTAL = 10+1= 11ml
[CH3COOH] =
0.9056
11
= 0.0823
[CH3COONa] =
0.0982
11
=8.9272 X 10 -3
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[8.9272𝑋10−3]
[0.0823]
= 3.8

2. C) 2ml de NaOH
CH3COOH + NaOH = CH3COONa H2O
Meqinicial 1.0038 0.1964 ------- ------
Meqfinal 0.8074 ------ 0.1964 ------
VOLUMEN TOTAL = 10+2= 12ml
[CH3COOH] =
0.8074
12
= 0.0672 [CH3COONa] =
0.1964
12
= 0.01636
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.01636 ]
[0.0672]
= 4.14
D) 3ml de NaOH
CH3COOH + NaOH = CH3COONa H2O
Meqinicial 1.0038 0.2946 ------- ------
Meqfinal 0.7092 ------ 0.2946 ------
VOLUMEN TOTAL = 10+3= 13ml
[CH3COOH] =
0.7092
13
= 0.0545 [CH3COONa] =
0.2946
13
= 0.0226
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0226]
[0.0545]
= 4.37
E) 4ml de NaOH
CH3COOH + NaOH = CH3COONa H2O
Meqinicial 1.0038 0.3928 ------- ------
Meqfinal 0.6110 ------ 0.3928 ------
VOLUMEN TOTAL = 10+4= 14ml
[CH3COOH] =
0.6110
14
= 0.04364 [CH3COONa] =
0.3928
14
= 0.0280
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0280]
[0.04364]
= 4.6

3. F) 5ml de NaOH
CH3COOH + NaOH = CH3COONa H2O
Meqinicial 1.0038 0.4910 ------- ------
Meqfinal 0.5128 ------ 0.4910 ------
VOLUMEN TOTAL = 10+5= 15ml
[CH3COOH] =
0.5128
15
= 0.03418 [CH3COONa] =
0.4910
15
= 0.03273
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.03273]
[0.03418]
= 4.7
G) 6ml de NaOH
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 0.5892 ------- ------
Meqfinal 0.4146 ------ 0.5892 ------
VOLUMEN TOTAL = 10+6= 16ml
[CH3COOH] =
0.4146
16
= 0.02591 [CH3COONa] =
0.5892
16
= 0.0368
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0368 ]
[0.0259]
= 4.9
H) 7ml de NaOH
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 0.6874 ------- ------
Meqfinal 0.3164 ------ 0.6874 ------
VOLUMEN TOTAL = 10+7= 17ml
[CH3COOH] =
0.3164
17
= 0.0186 [CH3COONa] =
0.6874
17
= 0.0404
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0404 ]
[0.0186]
= 5.1

4. I) 7.5ml de NaOH
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 0.7365 ------- ------
Meqfinal 0.2673 ------ 0.7365 ------
VOLUMEN TOTAL = 10+7.5= 17.5ml
[CH3COOH] =
0.2673
17.5
= 0.0152 [CH3COONa] =
0.7365
17.5
= 0.0420
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0420 ]
[0.0152]
= 5.2
J) 8.5ml de NaOH
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 0.8347 ------- ------
Meqfinal 0.1691 ------ 0.8347 ------
VOLUMEN TOTAL = 10+8.5= 18.5ml
[CH3COOH] =
0.1691
18.5
= 9.1405𝑥10 − 3 [CH3COONa] =
0.8347
18.5
= 0.0451
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0451 ]
[9.1405𝑥10−3]
= 5.5
K) 9.5 ml de NaOH
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 0.9329 ------- ------
Meqfinal 0.0209 ------ 0.9329 ------
VOLUMEN TOTAL = 10+8.5= 19.5ml
[CH3COOH] =
0.0209
18.5
= 1.1297𝑥10 − 3 [CH3COONa] =
0.9329
19.5
= 0.0478
𝑝𝐻 = 𝑝𝑘𝑎 + 𝑙𝑜𝑔
[𝑠𝑎𝑙]
[𝑎𝑐𝑖𝑑𝑜 𝑑𝑒𝑏𝑖𝑙]
p𝐻 = 4.7569 + 𝑙𝑜𝑔
[0.0478 ]
[1.1297𝑥10−3]
= 6.4

5. L) 10.22ml de NaOH en el punto de equivalencia
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 1.0038 ------- ------
Meqfinal 0 ------ 1.0038 ------
VOLUMEN TOTAL = 10+10.22= 20.22ml
Contenido del matraz: CH3COONa = pH de la sal alcalina
[CH3COONa] =
1.0038
20.22
= 0.09821
[𝑂𝐻] = √(kw/ka)(sal alcalina) =
[𝑂𝐻] = √(1x10 − 14/1.75x10 − 5)(0.09821) = 7.4913x10 − 6
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
7.4913x10 − 6
= 1.3348𝑥10 − 9
𝑝𝐻 = − log(1.3348𝑥10 − 9) = 8.9
M) 11ml de NaOH despues del punto de equivalencia
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 1.0802 ------- ------
Meqfinal 0 0.0764 1.0038 ------
VOLUMEN TOTAL = 10+11= 21ml
Contenido en el matraz: NaOH , pH del NaOH
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.0764
21
= 3.6380𝑥10 − 3
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
3.6380𝑥10 − 3
= 2.7487𝑥10 − 12
𝑝𝐻 = − log(2.7487𝑥10 − 12) = 11.7

6. N) 12ml de NaOH despues del punto de equivalencia
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 1.1784 ------- ------
Meqfinal 0 0.1746 1.0038 ------
VOLUMEN TOTAL = 10+12= 22ml
Contenido en el matraz: NaOH , pH del NaOH
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.1746
22
= 7.9363𝑥10 − 3
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
7.9363𝑥10 − 3
= 1.2600𝑥10 − 12
𝑝𝐻 = − log(1.2600𝑥10 − 12) = 11.9
O) 13ml de NaOH despues del punto de equivalencia
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 1.2766 ------- ------
Meqfinal 0 0.2728 1.0038 ------
VOLUMEN TOTAL = 10+13= 23ml
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.2728
23
= 0.01186
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
0.01186
= 8.4310𝑥10 − 13
𝑝𝐻 = − log(8.4310𝑥10 − 13) = 12
P) 14ml de NaOH despues del punto de equivalencia
CH3COOH + NaOH = CH3COONa + H2O
Meqinicial 1.0038 1.3748 ------- ------
Meqfinal 0 0.3710 1.0038 ------
VOLUMEN TOTAL = 10+14= 24ml
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.3710
24
= 0.01545
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
0.01545
= 6.4690𝑥10 − 13
𝑝𝐻 = − log(6.4690𝑥10 − 13) = 12.2

7. Curvas de titulación teoricas de
Ácido débil más Base fuerte
CH3
COOH + NaOH = CH3
COONA +H2
O
mL NaOH pH dml d pH/mL D2mL D2pH/d2mL
0 2.9 0 0 0 0
1 3.8 0.5 1.8 0.25 7.2
2 4.14 1.5 0.34 1 -1.94666667
3 4.37 2.5 0.23 2 -0.11
4 4.6 3.5 0.23 3 -8.8818E-16
5 4.7 4.5 0.1 4 -0.13
6 4.9 5.5 0.2 5 0.1
7 5.1 6.5 0.2 6 -8.8818E-16
7.5 5.2 7.25 0.133333333 6.875 -0.07619048
8.5 5.5 8 0.4 7.625 0.35555556
9.5 6.4 9 0.9 8.5 0.57142857
10.22 8.9 9.86 2.906976744 9.43 2.15803951
11 11.7 10.61 3.733333333 10.235 1.02652992
12 11.9 11.5 0.224719101 11.055 -4.27879784
13 12 12.5 0.1 12 -0.13197788
14 12.2 13.5 0.2 13 0.1
0
2
4
6
8
10
12
14
0 2 4 6 8 10 12 14 16
PH

8. Curvas de titulación teóricas de
Ácido débil más Base fuerte
CH3
COOH + NaOH = CH3
COONA +H2
O
0
0.5
1
1.5
2
2.5
3
3.5
4
0 2 4 6 8 10 12 14 16
ml
pH/mL
Primera derivada
-6
-4
-2
0
2
4
6
8
0 2 4 6 8 10 12 14
mL
pH/mL
Segunda Derivada

9. Curvas de titulación prácticas de
Ácido débil más Base fuerte
CH3
COOH + NaOH = CH3
COONA +H2
O
mL NaOH pH dml dph/mL D2ml D2pH/d2mL
0 2.91 0 0 0 0
2 3.87 1 0.48 0 0
4 4.39 3 0.26 2 -0.11
5 4.61 4.5 0.22 3.75 -0.026666
6 4.79 5.5 0.18 5 -0.04
7 4.99 6.5 0.2 6 0.02
7.5 5.1 7.25 0.22 6.875 0.026666
7.6 5.15 7.55 0.5 7.4 1.12
7.9 5.19 7.75 0.1333 7.65 -1.83335
8.1 5.25 8 0.3 7.875 -4.132
8.4 5.32 8.25 0.23333 8.125 -0.2668
8.7 5.4 8.55 0.26666 8.4 0.111
8.9 5.46 8.8 0.3 8.675 0.1336
9.1 5.85 9 1.95 8.9 8.25
9.3 6.04 9.2 0.95 9.1 -5
9.4 6.65 9.35 6.1 9.275 34.33333
9.5 7.08 9.45 4.3 9.4 -18
10 10.48 9.75 6.8 9.6 8.3333
10.5 11.08 10.25 1.2 10 -11.2
11 11.35 10.75 0.54 10.5 1.32
12 11.54 11.5 0.19 11.125 -0.46666

10. Curvas de titulación prácticas de
Ácido débil más Base fuerte
CH3
COOH + NaOH = CH3
COONA +H2
O