Calculate the pH of a solution prepared by mixing 10.00mL of 0.50M NaOH and 20.00mL of 0.50M benzoic acid solution. (Benzoic acid is monoprotic; its ionization constant is 6.7x10^-5).
I'm almost positive that I posted a response to this a couple of days ago. I won't go through it again. If you aren't the one I posted for and you still are confused on what to do about this AND the HOCl problem, post a new question at the top of the board and explain what you understand and what you don't understand. Give explicit details. I can spend an hour working these two, if you are just checking your answers, and I don't want to waste my time.
To calculate the pH of a solution prepared by mixing NaOH and benzoic acid, we need to understand the chemical reaction that occurs when they are mixed.
When NaOH (sodium hydroxide) and benzoic acid react, they form sodium benzoate and water. The balanced chemical equation for this reaction is:
C6H5COOH (benzoic acid) + NaOH (sodium hydroxide) → C6H5COONa (sodium benzoate) + H2O (water)
Since benzoic acid is monoprotic (has one ionizable hydrogen atom), it fully ionizes in water, so we can assume that all benzoic acid molecules will lose their hydrogen ion (H+) and form benzoate ions (C6H5COO-).
Now, let's determine the moles of NaOH and benzoic acid used in the solution.
Moles of NaOH = concentration of NaOH × volume of NaOH in liters
Moles of NaOH = 0.5 M × (10.00 mL ÷ 1000 mL/ L) = 0.005 mol
Moles of benzoic acid = concentration of benzoic acid × volume of benzoic acid in liters
Moles of benzoic acid = 0.5 M × (20.00 mL ÷ 1000 mL/ L) = 0.01 mol
Now, we need to determine the limiting reagent, which is the reactant that is fully consumed in a reaction and determines the amount of product formed.
By comparing the moles of NaOH (0.005 mol) and benzoic acid (0.01 mol), we see that NaOH is the limiting reagent because it is in stoichiometric excess compared to the benzoic acid.
Since the limiting reagent is NaOH, it will be fully consumed, and the remaining benzoic acid will react with the sodium hydroxide to form sodium benzoate.
To determine the concentration of sodium benzoate formed, we can use the stoichiometry of the balanced chemical equation. From the reaction equation, we see that 1 mole of benzoic acid reacts with 1 mole of NaOH to form 1 mole of sodium benzoate. Therefore, the concentration of sodium benzoate will be the same as the concentration of excess benzoic acid, which is 0.01 M.
Now, we need to calculate the concentration of H+ ions (Hydrogen ions) in the solution. Since benzoic acid is a weak acid, it does not fully dissociate in water. Instead, it undergoes partial ionization. The ionization constant (Ka) of benzoic acid is given as 6.7x10^-5.
The equation for the ionization of benzoic acid is as follows:
C6H5COOH (benzoic acid) ⇌ C6H5COO- (benzoate ion) + H+ (Hydrogen ion)
The ionization constant expression for benzoic acid can be written as:
Ka = [C6H5COO-] × [H+] / [C6H5COOH]
Since the concentration of benzoic acid is equal to the concentration of sodium benzoate (0.01 M), we can rewrite the Ka expression as:
Ka = [C6H5COO-] × [H+] / 0.01 M
Given that the Ka value is 6.7x10^-5, we can rearrange the equation and solve for [H+]:
[H+] = (Ka × 0.01 M) / [C6H5COO-]
Substituting the Ka value and the concentration of sodium benzoate into the equation, we can calculate [H+]:
[H+] = (6.7x10^-5 × 0.01 M) / 0.01 M
[H+] = 6.7x10^-5 M
To find the pH of the solution, we can use the pH formula:
pH = -log[H+]
Substituting the [H+] value into the equation, we get:
pH = -log(6.7x10^-5)
pH ≈ 4.17
Therefore, the pH of the solution prepared by mixing 10.00 mL of 0.50 M NaOH and 20.00 mL of 0.50 M benzoic acid is approximately 4.17.