what is the ph of a solution that results from adding 30 mL of .015M KOH to 50 mL of .015 M benzoic acid?
Benzoic acid = HB
millimols HB = 50 x 0.015 = 0.750
mmols KOH = 30 x 0.015 = 0.45
............HB + KOH ==>KB + H2O
initial...0.75...0.45....0....0
change...-0.45..-0.45.+0.45..+0.45
equil.....0.30....0.....0.45..0.45
So what you end up with is benzoic acid (0.30 mmols--a weak acid) and potassium benzoata (0.45 mmols--the salt of a weak base) which is a buffered solution.
Use the Henderson-Hasselbalch equation to solve for pH.
To determine the pH of a solution resulting from the addition of KOH to benzoic acid, we need to consider the reaction between these two chemicals.
The balanced chemical equation for the reaction between KOH (potassium hydroxide) and benzoic acid is:
C6H5COOH + KOH → C6H5COOK + H2O
From the equation, we can see that one mole of benzoic acid reacts with one mole of KOH to produce one mole of potassium benzoate (C6H5COOK) and one mole of water (H2O).
Given that the volume of KOH solution added is 30 mL and the concentration is 0.015 M, we can calculate the moles of KOH added using the formula:
moles of KOH = volume (in liters) × concentration (in M)
= 0.030 L × 0.015 mol/L
= 0.00045 moles
Since the reaction is stoichiometric (1:1) between KOH and benzoic acid, the number of moles of benzoic acid consumed will also be 0.00045 moles.
To find the concentration of benzoic acid in the final solution, we need to calculate the total volume of the solution, which is the sum of the initial volumes of KOH and benzoic acid:
total volume = volume of KOH + volume of benzoic acid
= 30 mL + 50 mL
= 80 mL
= 0.080 L
Now, we can calculate the concentration of benzoic acid using the formula:
final concentration = moles of benzoic acid / total volume (in liters)
= 0.00045 mol / 0.080 L
= 0.005625 M
To determine the pH, we need to calculate the pKa of benzoic acid, which is the logarithmic form of the acid dissociation constant (Ka). The pKa for benzoic acid is approximately 4.20.
Since benzoic acid is a weak acid, it will partially dissociate into its conjugate base (benzoate ion) and release hydrogen ions (H+). At equilibrium, we can assume that the concentration of benzoic acid (undissociated form) is equal to the concentration of the benzoate ion (dissociated form).
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where:
pH = the pH of the solution
pKa = the pKa value of the acid (benzoic acid, in this case)
[A-] = concentration of the benzoate ion (C6H5COO-) in the solution
[HA] = concentration of the benzoic acid (C6H5COOH) in the solution
In our case, we know the pKa of benzoic acid is 4.20, and we have calculated the concentration of benzoic acid to be 0.005625 M. The concentration of the benzoate ion will be the same as the concentration of benzoic acid (as they are at equilibrium).
Plugging in the values:
pH = 4.20 + log (0.005625 / 0.005625)
= 4.20 + log 1
= 4.20
Therefore, the pH of the resulting solution is 4.20.
Note: This calculation assumes that the volumes of KOH and benzoic acid are additive and that no other reactions or complex equilibria are occurring.
To find the pH of the resulting solution, we need to consider the reaction between KOH (a strong base) and benzoic acid (a weak acid). The reaction between a strong base and a weak acid results in the formation of a salt and water.
First, let's find the number of moles of KOH and benzoic acid present in the solution:
Number of moles of KOH = volume (L) x concentration (mol/L)
= 0.030 L x 0.015 mol/L
= 0.00045 mol
Number of moles of benzoic acid = volume (L) x concentration (mol/L)
= 0.050 L x 0.015 mol/L
= 0.00075 mol
Since KOH is a strong base and benzoic acid is a weak acid, the KOH will react completely with benzoic acid, forming the salt potassium benzoate (KC7H5O2) and water (H2O).
The balanced equation for the reaction is:
C7H6O2 (benzoic acid) + KOH → KC7H5O2 (potassium benzoate) + H2O
From the balanced equation, we can see that 1 mole of benzoic acid reacts with 1 mole of KOH to form 1 mole of potassium benzoate. Therefore, the limiting reactant will be the one with fewer moles, which is KOH (0.00045 mol).
Since KOH is a strong base, it will fully dissociate in water, meaning that 0.00045 mol of KOH will form 0.00045 mol of OH- ions.
Now, let's calculate the concentration of OH- ions in the resulting solution:
Volume of resulting solution = volume of KOH + volume of benzoic acid
= 0.030 L + 0.050 L
= 0.080 L
Concentration of OH- ions = number of moles of OH- ions / volume of resulting solution
= 0.00045 mol / 0.080 L
= 0.005625 M
The concentration of OH- ions gives us the pOH of the solution. To find the pH, we can use the equation:
pH + pOH = 14
Therefore, the pH of the solution can be calculated as:
pH = 14 - pOH
= 14 - (-log[OH-])
= 14 - (-log(0.005625))
≈ 11.25
So, the pH of the solution that results from adding 30 mL of 0.015 M KOH to 50 mL of 0.015 M benzoic acid is approximately 11.25.