Question

Calcule el pH de las siguientes soluciones y clasifíquelas según dicho valor (ácido, neutro o base)
[H+] = 3.4 x 10-5 M
[H+] = 5.6 x 10-6 M
[H+] = 9.2 x 10-9 M
[H+] = 7.2 x 10-12 M
[OH-] = 6.3 x 10-6 M
[OH-] = 8.5 x 10-7 M
[OH-] = 1.8 x 10-8 M
[OH-] = 7.8 x 10-10 M

Actividad 2
Calcula la [H+] a dos cifras significativas de las soluciones que tienen los valores de pH y pOH siguientes
pH = 2.73
pH = 4.09
pH = 8.35
pOH = 3.75
pOH = 6.08
pOH = 9.10
pOH = 10.22

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Answer 1

Actividad 1

Para calcular el pH de las soluciones señaladas se utilizan las formulas indicadas (imagen adjunta)

Para clasificar una solución como ácida, neutra o base se considera lo siguiente:

pH menor a 7 es ácida

pH igual a 7 es neutra

pH mayor a 7 es base

Cálculos de pH

[H+] = 3.4 x 10-5 M

pH = - Log [H+] = - Log [3.4 x 10-5 M]  =

-Log 3.4 + Log 10-5 = -0.5 + 5 = 4.5

Solución ácida

[H+] = 5.6 x 10-6 M

pH = - Log [H+] = - Log [5.6 x 10-6 M]  =

-Log 5.6  + Log 10-6 = -0.7 + 6 = 5.3

Solución ácida

[H+] = 9.2 x 10-9 M

pH = - Log [H+] = - Log [9.2 x 10-9 M]  =

-Log 9.2 + Log 10-9 = -1 + 9 = 8

Solución básica

[H+] = 7.2 x 10-12 M

pH = - Log [H+] = - Log [7.2 x 10-12 M]  =

-Log 7.2 + Log 10-12 = -0,8 + 12 = 11.2

Solución básica

[OH-] = 6.3 x 10-6 M

pOH = - Log [OH+] = - Log [6.3 x 10-6 M]  =

-Log 6,3 + Log 10-6 = -0.8 + 6 = 5.2

pH + pOH = 14

pH = 14 – pOH = 14 – 5.2 = 8.8

Solución básica

[OH-] = 8.5 x 10-7 M

pOH = - Log [OH+] = - Log [8.5 x 10-7 M]  =

-Log 8.5 + Log 10-7 = -0,9 + 7 = 6.1

pH + pOH = 14

pH = 14 – pOH = 14 – 6.1 = 7.9

Solución básica

[OH-] = 1.8 x 10-8 M

pOH = - Log [OH+] = - Log [1.8 x 10-8 M]  =

-Log 1.8 + Log 10-8 = -0,2 + 8 = 7.2

pH + pOH = 14

pH = 14 – pOH = 14 – 7.2 = 6.8

Solución ácida

[OH-] = 7.8 x 10-10 M

pOH = - Log [OH+] = - Log [7.8 x 10-10 M]  =

-Log 7.8 + Log 10-10 = -0,9 + 10 = 9.1

pH + pOH = 14

pH = 14 – pOH = 14 – 9.1 = 4.9

Solución ácida

Actividad 2

Para calcular la [H+] a dos cifras significativas de las soluciones que tienen los valores de pH y pOH siguientes se utilizan las mismas formulas anteriores y se realiza el despeje correspondiente

Cálculos

pH = 2.73

pH = - Log [H+]

log [H+] = -pH = 2.73

[H+] = 10-2.73 =   5.37 x 102 M

pH = 4.09

pH = - Log [H+]

log [H+] = -pH = 4.09

[H+] = 10-4.09 =   1.23 x 104 M

pH = 8.35

pH = - Log [H+]

log [H+] = -pH = 8.35

[H+] = 10-8.35 =   2,24 x 108 M

pOH = 3.75

pH + pOH = 14

PH = 14 – pOH = 14 – 3.75 = 10.25

pH = - Log [H+]

log [H+] = -pH = 10.25

[H+] = 10-10.25 =  1.78 X 1010 M

pOH = 6.08

pH + pOH = 14

pH = 14 – pOH = 14 – 6.08 = 7.92

pH = - Log [H+]

log [H+] = -pH = 7.92

[H+] = 10-7.92 =   8.32 x 107 M

pOH = 9.10

pH + pOH = 14

PH = 14 – pOH = 14 – 9.10 = 4.90

pH = - Log [H+]

log [H+] = -pH = 4.90

[H+] = 10-4.90 = 7.94 x 104  M

pOH = 10.22

pH + pOH = 14

PH = 14 – pOH = 14 – 10.22= 3.78

pH = - Log [H+]

log [H+] = -pH = 3.78

[H+] = 10-3.78 =   6.02x103 M

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