quimica














pH y pOH

1.- Calcule el pH y diga si son ácidas o bases las siguientes disoluciones:

pH = log [H+]

a) [H+] = 2 x 10-3
pH = log [2 x 10-3 ] = 2.69 baseacida.

b) [H+] = 5.2 x 10-4
pH = log [2 x 10-3 ] = 3.28 base acida.

c) [H+] = 8 x 10-6
pH = log [8 x 10-6] = 5.09 base acida.

d) [H+] = 4.5 x 10-5
pH = log [4.5 x 10-5] = 4.34base acida.


2.- Calcule el pOH de las siguientes disoluciones:
-log [-(OH)]

a) con [-(OH)] = 2.1 x 10-2
-log [-(2.1x10-2)] = 1.67 base acida
b) con [-(OH)] = 2 x 10-4
-log [-(2x 10-4)] = 3.69 base acida
c) con [-(OH)] = 7.8 x 10-10
-log [-(7.8 x 10-10)] = 9.10 base base
d) con [-(OH)] = 4.8 x 10-6
-log [-(4.8 x 10-6 )] = 5.31 baseacida


3.- Calcule el pH y diga si son ácidas o bases las siguientes disoluciones:

pH + Poh = 14
a) con [-(OH)] =
. pOH = - log [-(10-12)] = 12.00
pH = 14 – Poh = 14 - 12.00 =2.00 base acida
b) con [-(OH)] = 3 x 10-11
. pOH = - log [-(3 x 10-11)] = 10.52
pH = 14 – Poh = 14 – 10.52 = .47 base acida
c) con [-(OH)] = 2.4 x 10-10
. pOH = - log [-(2.4 x10-10)] = 9.62
pH = 14 – Poh = 14 – 9.62 = .4.38 base acida
d) con [-(OH)] = 6.2 x 10-12
pOH = - log [-(6.2 x 10-12)] = 11.20
pH = 14 – Poh = 14 – 11.20 = .2.79 base acida


4.- Calcule el pOHde las siguientes disoluciones:
a) Con [H+] = 2.1 x 10-2
pH = log [2.1 x 10-2] = 1.67
ph + poh = 14
1.67 + pOH = 14
pOH = 14 - 1.67
pOH = 12.33 base base
b) Con [H+] = 8.5x 10-9
pH = log [8.5 x 10-9] = 8.07
ph + poh = 14
1.67 + pOH = 14
pOH = 14 – 8.07
pOH = 5.93 base acida

c) Con [H+] = 3.75 x 10-4
pH = log 3.75 x 10-4] = 3.42
ph + poh =14
1.67 + pOH = 14
pOH = 14 – 3.42
pOH = 10.58 base base

d) Con [H+] = 1.5 x 10-6
pH = log [1.5 x 10-6] = 5.82
ph + poh = 14
1.67 + pOH = 14
pOH = 14 – 5.82
pOH = 8.18 base base...