The document describes the theoretical and experimental titration curves of a strong acid (HCl) with a strong base (NaOH). Theoretical calculations are shown for pH as a function of added NaOH volume from 0-12 mL. Experimentally, pH is measured at increasing volumes from 0-9.3 mL NaOH. Both curves show pH increasing sharply near 8 mL NaOH due to the equivalence point of the reaction.

1. Cálculos teóricos de la valoración de
Ácido fuerte más Base fuerte
HCl + NaOH = NaCl +H2
O
NaOH= 0.1044 N
𝑁1𝑉1 = 𝑁2𝑉2 𝑉1 =
(0.0915)(10)
0.1044
= 8.76𝑚𝑙
VOL=10ml
HCl= 0.0915 N
8.76 ml punto de equivalencia teórico.
A) 0ml de NaOH
En el matraz solo hay HCl
[𝐻] = [𝐻𝐶𝑙] = 0.0915
𝑝𝐻 = −𝑙𝑜𝑔[𝐻]
𝑝𝐻 = − log(0.0915) = 1.03 pH inicial
B) 1ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.1044 ------- ------
Meqfinal 0.8106 ------ 0.1044 ------
VOLUMEN TOTAL = 10+1= 11ml
Contenido en el matraz: HCl + NaCl; pH del HCl
[𝐻] = [𝐻𝐶𝑙] =
0.8106
11
= 0.0736
𝑝𝐻 = − log(0.0736) = 1.13
C) 2ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.2088 ------- ------
Meqfinal 0.7062 ------ 0.2088 ------
VOLUMEN TOTAL = 10+2= 12ml
[𝐻] = [𝐻𝐶𝑙] =
0.7062
12
= 0.0588
𝑝𝐻 = − log(0.7062) = 1.23

2. D) 4ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.4176 ------- ------
Meqfinal 0.4974 ------ 0.4176 ------
VOLUMEN TOTAL = 10+4= 14ml
[𝐻] = [𝐻𝐶𝑙] =
0.4974
14
= 0.0355
𝑝𝐻 = − log(0.0355) = 1.44
E) 6ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.6264 ------- ------
Meqfinal 0.2886 ------ 0.6264 ------
VOLUMEN TOTAL = 10+6= 16ml
[𝐻] = [𝐻𝐶𝑙] =
0.2886
12
= 0.02405
𝑝𝐻 = − log(0.02405) = 1.6
F) 7ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.7308 ------- ------
Meqfinal 0.1842 ------ 0.7308 ------
VOLUMEN TOTAL = 10+7= 17ml
[𝐻] = [𝐻𝐶𝑙] =
0.1842
17
= 0.0108
𝑝𝐻 = − log(0.0108) = 2.0
G) 8ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.8352 ------- ------
Meqfinal 0.0798 ------ 0.8352 ------
VOLUMEN TOTAL = 10+8= 18ml
[𝐻] = [𝐻𝐶𝑙] =
0.0798
18
= 4.4333𝑥10 − 3
𝑝𝐻 = − log(4.4333𝑥10 − 3) = 2.35

3. H) 8.5ml de NaOH
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.8874 ------- ------
Meqfinal 0.0276 ------ 0.8874 ------
VOLUMEN TOTAL = 10+8.5= 18.5ml
[𝐻] = [𝐻𝐶𝑙] =
0.0276
18.5
= 1.4918𝑥10 − 3
𝑝𝐻 = − log(1.4918𝑥10 − 3) = 2.82
H) 8.76ml de NaOH en el punto de equivalencia
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.9150 ------- ------
Meqfinal 0 ------ 0.9150 ------
VOLUMEN TOTAL = 10+8.76= 18.76ml
Contenido en el matraz: NaCl; pH de sal neutra
En el punto de equivalencia [H] =[OH]
Kw=[H] =[OH]= 1x10-14
Kw=[H] =[H]
Kw=[H]2
=1x10-14
[𝐻] = √𝑘𝑤 = √1𝑥10 − 14 = 1𝑥10 − 7
𝑝𝐻 = − log(1𝑥10 − 7) = 7
I) 9ml de NaOH después del punto de equivalencia
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 0.9396 ------- ------
Meqfinal 0 0.0246 0.9150 ------
VOLUMEN TOTAL = 10+9= 19ml
Contenido en el matraz: NaOH; pH del NaOH
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.0246
19
= 1.2947𝑋10 − 3
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
1.2947𝑋10 − 3
= 7.7235𝑋10 − 12
𝑝𝐻 = − log(7.7235𝑋10 − 12) = 11.11

4. J) 10ml de NaOH después del punto de equivalencia
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 1.044 ------- ------
Meqfinal 0 0.1290 0.9150 ------
VOLUMEN TOTAL = 10+10= 20ml
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.1290
20
= 6.450𝑥10 − 3
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
6.450𝑥10 − 3
= 1.5503𝑥10 − 12
𝑝𝐻 = − log(1.5503𝑥10 − 12) = 11.8
k) 11ml de NaOH después del punto de equivalencia
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 1.1484 ------- ------
Meqfinal 0 0.2334 1.1484 ------
VOLUMEN TOTAL = 10+11= 21ml
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.2334
21
= 0.0111
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
0.0111
= 8.9974𝑥10 − 13
𝑝𝐻 = − log(8.9974𝑥10 − 13) = 12.04
L) 12ml de NaOH después del punto de equivalencia
HCl + NaOH = NaCl + H2O
Meqinicial 0.9150 1.2528 ------- ------
Meqfinal 0 0.3378 1.2528 ------
VOLUMEN TOTAL = 10+12= 22ml
[𝑂𝐻] = [𝑁𝑎𝑂𝐻] =
0.3378
22
= 0.01535
[𝐻] =
𝑘𝑤
[𝑂𝐻]
=
1𝑥10 − 14
0.01535
= 6.5127𝑥10 − 13
𝑝𝐻 = − log(6.5127𝑥10 − 13) = 12.2

5. Curvas de titulación teóricas de
Ácido fuerte más Base fuerte
HCl + NaOH = NaCl +H2
O
mL NaOH pH dml dph/mL d2ml d2ph/d2ml
0 2.9 0 0 0 0
1 3.8 0.5 0 0 0
2 4.14 1.5 0.34 1 0
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
MLNAOH
PH
PH

6. Curvas de titulación teóricas de
Ácido fuerte más Base fuerte
HCl + NaOH = NaCl +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
dpH/mL
Primera Derivada
-5
-4
-3
-2
-1
0
1
2
3
0 2 4 6 8 10 12 14
mL
pH/mL
Segunda derivada

7. Curvas de titulación prácticas de
Ácido fuerte más Base fuerte
HCl + NaOH = NaCl +H2
O
mL NaOH pH dml d pH/mL dmL2 dpH/mL
0 1.66 0 0 0 0
1 1.68 0.5 0.02 0 0
2 1.76 1.5 0.08 1 0.06
3 1.86 2.5 0.1 2 0.02
4 1.91 3.5 0.05 3 -0.05
5 2.02 4.5 0.11 4 0.06
6 2.15 5.5 0.13 5 0.02
7 2.41 6.5 0.26 6 0.13
7.5 2.52 7.25 0.22 6.875 -0.05333
8 2.84 7.75 0.64 7.5 0.84
8.5 4.05 8.25 4.42 8 7.56
8.6 5.46 8.55 14.1 8.4 32.26666
8.7 7.05 8.65 15.9 8.6 18
8.8 9.24 8.75 21.9 8.7 60
9.3 10.91 9.05 3.34 8.9 -61.86666

8. Curvas de titulación prácticas de
Ácido fuerte más Base fuerte
HCl + NaOH = NaCl +H2
O