The disssociation constant of benzoic acid is 1x10 power -4 . the pH of 0.01M solution of its sodium salt will be......?
To determine the pH of a solution of sodium benzoate (the sodium salt of benzoic acid), you need to consider the dissociation of the sodium benzoate in water.
First, let's understand the dissociation reaction of sodium benzoate:
C6H5COO- (sodium benzoate) + H2O ↔ C6H5COOH (benzoic acid) + OH- (hydroxide ion)
Since sodium benzoate completely dissociates in water, we can assume that 0.01 M of sodium benzoate will produce the same concentration of benzoic acid and hydroxide ion.
Now, let's write the expression for the dissociation constant (Ka) of benzoic acid:
Ka = [C6H5COOH] / [C6H5COO-] * [OH-]
However, we know that [C6H5COO-] = [C6H5COOH]. Therefore, we can rewrite the expression as:
Ka = [OH-]
We are given the dissociation constant (Ka) as 1 x 10^-4, which represents the concentration of OH- in the solution.
To calculate the pOH, we can take the negative logarithm of the concentration of OH-:
pOH = -log10([OH-])
Therefore, pOH = -log10(1 x 10^-4) = 4.
Finally, to find the pH, we can use the relation pH + pOH = 14:
pH = 14 - pOH = 14 - 4 = 10.
Therefore, the pH of the 0.01 M solution of sodium benzoate will be 10.