500.0 mL of 0.130 M NaOH is added to 625 mL of 0.250 M weak acid (Ka=5.17×10−5). What is the pH of the resulting buffer?
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
(NaOH) = 0.130 x (500/1125) = ?= approx 0.06
(HA) = 0.250 x (625/1125) = ? = approx 0.14
.......... HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
I..........0.14.......0..................0..........0
add,,,,,,,,,,,,,,,,..0.06..............................
C.....-0.06.....-0.06...........0.06........0.06
E.........?...........0...............0.06.......0.06
Note: all of these numbers are estimates.
Plug the E line into the HH equation and solve or pH.
Post your work if you get stuck.
To find the pH of the resulting buffer solution, we need to calculate the concentration of the acid and its conjugate base after the NaOH is added.
First, let's determine the moles of acid and base used:
Moles of NaOH = concentration x volume
= 0.130 M x 500.0 mL
= 0.065 mol
Moles of weak acid = concentration x volume
= 0.250 M x 625 mL
= 0.15625 mol
Since the acid and base react in a 1:1 ratio, the amount of excess acid or base is the difference in moles between the acid and base used:
Excess moles of acid = moles of weak acid - moles of NaOH
= 0.15625 mol - 0.065 mol
= 0.09125 mol
The concentration of the weak acid after the reaction is the excess moles divided by the total volume:
Concentration of weak acid = moles / total volume
= 0.09125 mol / (500 mL + 625 mL)
= 0.09125 mol / 1.125 L
= 0.081 ml
The concentration of the conjugate base (A-) is the same as the concentration of NaOH used:
Concentration of A- = concentration of NaOH
= 0.130 M
Now, we can use the Henderson-Hasselbalch equation to find the pH of the resulting buffer:
pH = pKa + log([A-] / [HA])
Since the pKa is given and is equal to -log(Ka), we can substitute the values into the equation:
pH = -log(5.17x10^-5) + log(0.130 / 0.081)
Simplifying:
pH = -log(5.17x10^-5) + log(1.6049)
Using the properties of logarithms:
pH = 4.287 + 0.204
Adding the numbers:
pH = 4.491
Therefore, the pH of the resulting buffer is approximately 4.491.
To find the pH of the resulting buffer, we need to consider the equilibrium between the weak acid (HA) and its conjugate base (A-). The equation you provided represents the dissociation of HA into H2O and A-.
Since we are adding NaOH (a strong base) to the weak acid, it will react completely with the weak acid to form water and the conjugate base. This reaction consumes all the OH- ions from NaOH. As a result, the concentration of OH- in the final buffer solution will depend solely on the concentration of A-.
To solve this problem, we'll follow these steps:
1. Calculate the moles of the weak acid and its conjugate base in the solution.
2. Calculate the moles of OH- ions produced by the complete reaction of NaOH with the weak acid.
3. Determine the new concentrations of the weak acid and its conjugate base after the reaction.
4. Use the Henderson-Hasselbalch equation to calculate the pH of the resulting buffer.
Let's go through each step:
Step 1: Calculate the moles of the weak acid and its conjugate base
Molarity (M) is defined as moles of solute per liter of solution. Since we have the volume, we can calculate the moles of the weak acid and its conjugate base:
moles of HA = volume of weak acid solution (L) * concentration of HA (M)
= 625 mL * 0.250 M (convert mL to L: divide by 1000)
= 0.625 L * 0.250 M
= 0.15625 moles
moles of A- = volume of NaOH solution (L) * concentration of NaOH (M)
= 500 mL * 0.130 M (convert mL to L: divide by 1000)
= 0.500 L * 0.130 M
= 0.065 moles
Step 2: Calculate the moles of OH- ions produced
Since NaOH is a strong base, it will react completely with the weak acid. The balanced equation you provided shows that 1 mole of HA reacts with 1 mole of OH- to form 1 mole of A-, so the moles of OH- ions produced is equal to the moles of A-:
moles of OH- = 0.065 moles
Step 3: Determine the new concentrations of HA and A-
To determine the new concentrations after the reaction, we need to consider the final volume of the solution. The total volume is the sum of the volumes of the weak acid and NaOH solutions:
total volume = volume of weak acid solution + volume of NaOH solution
= 625 mL + 500 mL
= 1125 mL (convert mL to L: divide by 1000)
= 1.125 L
The new concentration of HA is the moles of HA divided by the total volume:
new [HA] = moles of HA / total volume
= 0.15625 moles / 1.125 L
= 0.139 M
The new concentration of A- is the moles of A- divided by the total volume:
new [A-] = moles of A- / total volume
= 0.065 moles / 1.125 L
= 0.058 M
Step 4: Calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the weak acid and its conjugate base:
pH = pKa + log([A-] / [HA])
In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka).
pH = -log(Ka) + log([A-] / [HA])
Now we can substitute the known values:
pH = -log(5.17×10^(-5)) + log(0.058 M / 0.139 M)
pH = -(-4.29) + log(0.418)
pH = 4.29 + (-0.378)
pH = 3.91
Therefore, the pH of the resulting buffer is approximately 3.91.