calculate the pH of 5x10^-2 M of benzoic acid with pKa= 4.19
Refer to you ASA problem above. This one is done th same way.
To calculate the pH of a solution, we need to determine the concentration of H+ ions in the solution.
For benzoic acid (C6H5COOH), it can partially dissociate into H+ ions and its conjugate base, C6H5COO-. The dissociation reaction is as follows:
C6H5COOH ⇌ H+ + C6H5COO-
The dissociation constant for benzoic acid is referred to as pKa and is given as 4.19 in this case. The pKa is the logarithmic form of the equilibrium constant (Ka) for the dissociation reaction.
To calculate the pH, we need to consider the relative amounts of the acidic form (benzoic acid) and the basic form (C6H5COO-) in the solution. The pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([C6H5COO-]/[C6H5COOH])
In this case, we are given the concentration of benzoic acid as 5x10^-2 M. Since the acid partially dissociates, the concentration of benzoic acid ([C6H5COOH]) will decrease by some amount, while the concentration of the conjugate base ([C6H5COO-]) will increase by the same amount.
Let's substitute the given values into the Henderson-Hasselbalch equation:
pH = 4.19 + log([C6H5COO-]/[C6H5COOH])
Now, we need to determine the ratio of [C6H5COO-]/[C6H5COOH].
Since the dissociation reaction is 1:1, the ratio of the conjugate base to the acid is equal to the ratio of their concentrations. Therefore, the ratio is ([C6H5COO-]/[C6H5COOH]) = x/x, where x is the amount (in moles or M) by which the benzoic acid has dissociated.
Now, we can rewrite the Henderson-Hasselbalch equation:
pH = 4.19 + log(x/x)
Simplifying the equation:
pH = 4.19 + log(1)
Since log(1) = 0, we can simplify further:
pH = 4.19 + 0
Finally, the solution is:
pH = 4.19
Therefore, the pH of a 5x10^-2 M benzoic acid solution with a pKa of 4.19 is 4.19.