A 1.0L buffer solution contains 0.100 mol of HC2H3O2
and 0.100 mol of NaC2H3O2
The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl
to the original buffer.
I worked this buffer problem for someone yesterday, perhaps you, that asked for the pH after NaOH was added. I left instructions for the HCl addition. Now you want me to do it for you.
HC2H3O2 is HAc in shorthand notation and Ac^- is acetate ion.
pKa = -log Ka = 4.74
The Henderson-Hasselbalch equation is
pH = pKa + log (Ac^-)/(HAc)
You are adding 10 mL of 1 M HCl which is 0.01 x 1 = 0.01 mols.
At equilibrium you have
.........................Ac^- + H^+ ==> HAc
Initial mols........0.1......0.............0.1
add............................0.01..........................
change...........-0.01..-0.01.......... 0.01
equilibrium......0.09.......0..............0.11
Plug the equilibrium line into the HH equation and solve for pH.
Just a note here because some profs are picky. You will note that the HH equation uses CONCENTRATION OF Ac^- and HAc. I have used mols above. If you want to be proper, you can take the moles Ac and mols HAc and convert them to concn in mols/L. (Ac^-) = 0.09/1.01 L = ? and (HAc) = 0.115/1.01 L and use those instead of mols in the HH equation/ However, since the volume is the same in both cases (1.01 L), that number cancels. You get the same answer but picky profs will count off if moles are used instead of concn because concn is in the formula. Post your work if you get stuck. BTW, I was a picky prof and I gave only partial credit if moles were used without the volume. Some clever students who didn't want to go through the extra calculation to calculate M would show it as
pH = 4.74 + log (0.09/V)/(0.11/V). That I had to recognize that 0.09/V was the concentration. Those clever students received full credit and didn't go through the extra work either.
To calculate the pH of the solution after adding the HCl to the buffer, we need to consider the reaction that occurs between HCl and HC2H3O2.
Step 1: Determine the initial concentrations of HC2H3O2 and NaC2H3O2 in the buffer solution.
In this case, we have 0.100 mol of HC2H3O2 in 1.0L of solution, so the initial concentration of HC2H3O2 is 0.100 M.
Similarly, we have 0.100 mol of NaC2H3O2 in 1.0L of solution, so the initial concentration of NaC2H3O2 is also 0.100 M.
Step 2: Write the balanced chemical equation for the reaction between HCl and HC2H3O2.
HC2H3O2 + HCl ⇌ H3O+ + C2H3O2- (acetic acid + hydrochloric acid ⇌ hydronium ion + acetate ion)
Step 3: Determine the change in concentrations of HC2H3O2 and C2H3O2- due to the reaction with HCl.
Since HCl is a strong acid, it will completely dissociate in water, which means that all of the HCl will react with HC2H3O2 and form H3O+ and C2H3O2-. Therefore, the change in concentration of HC2H3O2 will be -0.010 M (10.0 mL of 1.00 M HCl added to the 1.0 L of 0.100 M HC2H3O2), and the change in concentration of C2H3O2- will also be -0.010 M.
Step 4: Calculate the equilibrium concentrations of HC2H3O2 and C2H3O2- after the reaction with HCl.
The initial concentration of HC2H3O2 is 0.100 M, and the change in concentration is -0.010 M, so the equilibrium concentration of HC2H3O2 is 0.090 M.
The initial concentration of C2H3O2- is 0.100 M, and the change in concentration is -0.010 M, so the equilibrium concentration of C2H3O2- is also 0.090 M.
Step 5: Calculate the concentration of H3O+ in the solution.
Since HC2H3O2 is a weak acid, we can assume that it only partially dissociates in water. According to the Ka value, the equilibrium expression for the dissociation of HC2H3O2 is:
Ka = [H3O+][C2H3O2-] / [HC2H3O2]
Given that the concentration of HC2H3O2 is 0.090 M and the concentration of C2H3O2- is 0.090 M, we can substitute these values and solve for [H3O+].
Step 6: Calculate the pOH and pH of the solution.
The pOH can be calculated using the equation:
pOH = -log10[OH-]
Since pOH + pH = 14 at 25°C, we can calculate the pH using the equation:
pH = 14 - pOH
By following these steps, you should be able to calculate the pH of the solution after adding the HCl to the buffer.
To calculate the pH of the solution upon the addition of HCl to the original buffer, we need to consider the reaction that occurs between the added HCl and the components of the buffer solution.
First, let's determine the initial concentrations of acetic acid (HC2H3O2) and acetate ion (C2H3O2-) in the buffer solution.
Given:
Volume of buffer solution = 1.0 L
Moles of HC2H3O2 = 0.100 mol
Moles of NaC2H3O2 = 0.100 mol
Since the moles of HC2H3O2 and NaC2H3O2 are equal, the buffer solution is prepared at the half-equivalence point of its conjugate acid-base pair. Thus, the initial concentrations of HC2H3O2 and C2H3O2- are both equal to:
Concentration = (moles of solute) / (volume of solution)
Concentration = 0.100 mol / 1.0 L
Concentration = 0.100 M
Now, let's consider the reaction between HCl and the components of the buffer solution:
HCl + HC2H3O2 -> H3O+ + C2H3O2-
In this reaction, HCl reacts with HC2H3O2 to form H3O+ (hydronium ion) and C2H3O2- (acetate ion).
We have 10.0 mL of 1.00 M HCl added to the original buffer solution.
To determine the final concentration of HC2H3O2 and C2H3O2- after the addition of HCl, we need to consider the reaction stoichiometry.
The moles of HCl added can be calculated as:
Moles of HCl = (concentration of HCl) * (volume of HCl)
Moles of HCl = 1.00 M * 0.010 L
Moles of HCl = 0.010 mol
Since the stoichiometric ratio between HCl and HC2H3O2 is 1:1, the moles of HC2H3O2 will decrease by 0.010 mol.
Therefore, the final moles of HC2H3O2 will be:
Final moles of HC2H3O2 = Initial moles of HC2H3O2 - Moles of HCl
Final moles of HC2H3O2 = 0.100 mol - 0.010 mol
Final moles of HC2H3O2 = 0.090 mol
Now, let's calculate the new concentration of HC2H3O2:
Concentration = (Final moles of solute) / (volume of solution)
Concentration = 0.090 mol / 1.0 L
Concentration = 0.090 M
Since the moles of NaC2H3O2 remain unchanged, its concentration remains 0.100 M.
With the new concentrations of HC2H3O2 and C2H3O2-, we can determine the pH of the solution using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
pKa is the negative logarithm of the acid dissociation constant Ka, which is given as 1.8×10−5.
Substituting the values into the Henderson-Hasselbalch equation:
pH = -log(1.8×10^-5) + log(0.100 M / 0.090 M)
Calculating the pH using a scientific calculator:
pH ≈ 4.75
Therefore, the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl to the original buffer is approximately 4.75.