# Why f is multiplied in solubility of salt constant

I understand till we get “f” And also why f is directly proportional to PH. But I don’t understand why f is multiplied in solubility product with solubility of salt of weak acid.

(Sorry for poor english) (couldn’t find link of pdf of this book online)

• What is this book? – andselisk Jan 12 '18 at 16:42
• 11th class ncert chemistry “acids bases and salts” – Fawad Jan 12 '18 at 16:49
• The notations are very poorly picked. Upright "S" stands for both sulfur element and solubility, brackets "[]" are used both for concentrations and to enclose variables, the font makes it impossible to distinguish between { and (, and so on (and this is only from what you photocopied). I've never seen this book before, but I'd rather get something better than this. – andselisk Jan 12 '18 at 18:54

Notice that $f$ is equal to the maximum concentration of $\ce{M}$, which is unchanged as part of the acid-base equilibrium. On the other hand, the pH changes the concentration of the counterion in proportion to $f$. Therefore, the second term in the solubility product is scaled accordingly.