Find the pH of 0.1M H3PO4 solution.
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its a tripotic acid with three different Ka values they are Ka1, Ka2 and Ka3.
Ka1 = H3PO4 = 7.5 * 10^-3
Ka2 = H2PO4^- = 6.2 * 10^-8
Ka3 = HPO4^2- = 4.8 * 10^-13
Ka1>>Ka2 and Ka3.
which says that majority of Hydronium is produced during step one so we can ignore step 2 and 3.
Since H3PO4 is weak acid it will not dissociate completely and we have to write the ICE table
H3PO4 + H2O ↔ H3O^+ + H2PO4^-
I 0.10M 0 0
C -x +x +x
E 0.10-x x x
Ka1 = ([H3O^+][H2PO4^-])/[H3PO4]
7.5*10^-3 = ([x][x])/[0.10-x]
Find x usind qudratic equation...
x = 0.024M
[H3O^+] = x = 0.024M
PH = -log[H3O^+]
PH = -log[0.024]
Ph = 1.619
The pH scale is a scale from 0 to 14 that indicates the acidity of a substance. The number 7 is in the middle and represents neutrality, in terms of qualifying a substance as an acid or a base. Substances that are lower than 7 are classified as acids, meaning they have increasing numbers of hydrogen ions (H+) when dissolved with water. Substances that are higher than 7 are classified as bases, meaning they have less hydrogen ions and more hydroxide ions (OH-) when dissolved with water.
To find the pH of a .1 M sample of H3PO4, understand the .1 M is the same as 1 x 10-1. If we take the logarithm of 1 x 10-1, that would be expressed as - (-1), which would be 1. The pH of this solution would be 1, which would qualify it as a very strong acid, providing many hydrogen ions upon dissolution with water. Th pH scale is very useful in indicating the acidity or basicity of substances.
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