Calculate the pH of a solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL
of 0.10 M benzoic acid (Ka = 6.5 x 10
-5
At pH 7.2, what is the ratio [HPO4 2-]/[H2PO4-]
To calculate the pH of the resulting solution, you need to consider the reaction between NaOH (a strong base) and benzoic acid (a weak acid).
First, let's determine the moles of benzoic acid and NaOH in the solution.
Moles of benzoic acid = concentration of benzoic acid x volume of benzoic acid
= 0.10 M x 0.050 L = 0.005 moles
Moles of NaOH = concentration of NaOH x volume of NaOH
= 0.10 M x 0.010 L = 0.001 moles
Now let's determine the amount of benzoic acid and NaOH that react with each other based on their stoichiometry. The balanced equation for the reaction is:
C6H5COOH + NaOH → C6H5COONa + H2O
From the equation, we can see that 1 mole of benzoic acid reacts with 1 mole of NaOH to produce 1 mole of C6H5COONa and 1 mole of water.
Since we have 0.005 moles of benzoic acid and 0.001 moles of NaOH, the limiting reactant is NaOH because we have fewer moles of NaOH.
Therefore, all of the 0.001 moles of NaOH will react with 0.001 moles of benzoic acid.
Now, let's calculate the moles of benzoic acid remaining after the reaction:
Moles of benzoic acid remaining = moles of benzoic acid initially - moles of benzoic acid that reacted
= 0.005 moles - 0.001 moles = 0.004 moles
To calculate the concentration of benzoic acid remaining, divide the moles remaining by the total volume of the solution:
Concentration of benzoic acid remaining = moles of benzoic acid remaining / total volume of solution
= 0.004 moles / (0.050 L + 0.010 L) = 0.004 moles / 0.060 L = 0.067 M
Now, let's use the equation for the dissociation of benzoic acid to calculate the concentration of H+ ions (protons) in the solution:
Ka = [H+][C6H5COO-] / [C6H5COOH]
[H+] = Ka x [C6H5COOH] / [C6H5COO-]
[H+] = (6.5 x 10^-5) x (0.067 M) / (0.067 M)
[H+] = 6.5 x 10^-5 M
The pH is determined by the negative logarithm (base 10) of the H+ ion concentration:
pH = -log[H+]
pH = -log(6.5 x 10^-5)
pH = 4.19
Therefore, the pH of the resulting solution is approximately 4.19.