What must be a concentration of benzoate ion in a 0.55M benzoic acid solution so that the ph is 5.00?
pH = pKa + log (benzoate)/(benzoic acid)
Look up pKa, You know benzoic acid and you know pH, solve for benzoate ion concentration.
To determine the concentration of benzoate ion that would result in a pH of 5.00 in a 0.55 M benzoic acid solution, we need to consider the dissociation of benzoic acid (C6H5COOH) into benzoate ion (C6H5COO-) and a hydrogen ion (H+).
The balanced chemical equation for the dissociation of benzoic acid is as follows:
C6H5COOH ⇌ C6H5COO- + H+
The dissociation of benzoic acid is given by the acid dissociation constant (Ka), which represents the ratio of the concentration of the products to the concentration of the reactants:
Ka = [C6H5COO-][H+] / [C6H5COOH]
Since we know the pH of the solution is 5.00, we can determine the concentration of hydrogen ions (H+) by using the formula:
pH = -log[H+]
Therefore, for a pH of 5.00:
[H+] = 10^(-pH) = 10^(-5.00) = 1.00 x 10^(-5) M
Now, let's assume 'x' is the concentration of benzoate ions ([C6H5COO-]). We can express the concentration of benzoic acid ([C6H5COOH]) in terms of 'x' as well:
[C6H5COOH] = 0.55 M - x
Now, substituting these values into the acid dissociation constant equation:
Ka = [C6H5COO-][H+] / [C6H5COOH]
Ka = (x)(1.00 x 10^(-5)) / (0.55 - x)
The Ka value for benzoic acid at 25°C is approximately 6.3 x 10^(-5) M. We can substitute this value into the equation:
6.3 x 10^(-5) = (x)(1.00 x 10^(-5)) / (0.55 - x)
Now, we can solve this equation for 'x' using algebraic manipulation:
6.3 x (0.55 - x) = x
3.465 - 6.3x = x
7.3x = 3.465
x = 0.475 M
Therefore, the concentration of benzoate ion in the 0.55 M benzoic acid solution, which would result in a pH of 5.00, is approximately 0.475 M.