100 mL 0.1 M sodium acetate solution is mixed with 50 mL 0.1 M acetic acid. If now 0.937 mL of 1 M sodium hydroxide is added to the buffer what is the change in pH? pKa (acetic acid) = 4.76
You may work this problem using the Henderson-Hasselbalch equation. Technically that uss concentration of base and acid but you may also use mols and the answer comes out the same. I use mols although concentration is supposed to be substituted.
millimols HAc = mL x M = 50 x 0.1 = 5
mmols NaAc = 100 x 0.1 = 10
mmols NaOH = 0.937 x 1 = 0.937
........HAc + OH^- ==> H2O + Ac^-
I.......5....0.........0.....10
add........0.973.................
C..-0.973..-0.973...........+.973
E...4.027....0............10.973
Substitute the E line into the H-H equation and solve for pH.
To find the change in pH after adding sodium hydroxide to the buffer, we need to consider the equilibrium reaction that occurs between acetic acid and sodium acetate:
CH3COOH + OH- ⇌ CH3COO- + H2O
First, let's calculate the initial concentrations of acetic acid (CH3COOH) and acetate ions (CH3COO-) in the buffer solution.
For the acetic acid solution:
Initial volume = 50 mL
Concentration = 0.1 M
Therefore, the initial amount (moles) of acetic acid in the solution can be calculated using:
Amount of acetic acid = Volume (L) x Concentration (M)
= (50 mL / 1000 mL/L) x 0.1 M
= 0.005 moles
For the sodium acetate solution:
Initial volume = 100 mL
Concentration = 0.1 M
Therefore, the initial amount (moles) of sodium acetate in the solution can be calculated using:
Amount of sodium acetate = Volume (L) x Concentration (M)
= (100 mL / 1000 mL/L) x 0.1 M
= 0.01 moles
Since acetic acid and acetate ions are in equal moles (1:1 ratio), the initial concentration of acetic acid and acetate ions in the mixture is 0.005 M.
The change in pH after adding sodium hydroxide can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([Acetate ions] / [Acetic acid])
Where pKa is the negative logarithm of the acid dissociation constant, in this case, 4.76.
Initially, the concentration of acetate ions ([Acetate ions]) and acetic acid ([Acetic acid]) are both 0.005 M.
pH_initial = 4.76 + log(0.005 / 0.005)
= 4.76
Now, let's find the final concentrations of acetate ions and acetic acid after adding 0.937 mL of 1 M sodium hydroxide to the buffer.
Since sodium hydroxide reacts with acetic acid in a 1:1 ratio, the moles of acetic acid neutralized by sodium hydroxide can be calculated using:
Moles of acetic acid neutralized = Volume (L) x Concentration (M)
= (0.937 mL / 1000 mL/L) x 1 M
= 0.000937 moles
Since acetic acid and acetate ions are in a 1:1 ratio, the moles of acetate ions formed will be the same as the moles of acetic acid neutralized.
Therefore, the final amount (moles) of acetate ions in the solution is:
Amount of acetate ions = Initial amount of sodium acetate + Moles of acetate formed
= 0.01 moles + 0.000937 moles
= 0.010937 moles
The final volume of the buffer solution is 150 mL (100 mL + 50 mL), or 0.150 L.
Therefore, the final concentrations of acetate ions and acetic acid in the buffer are:
[Acetate ions] = Amount of acetate ions / Volume (L)
= 0.010937 moles / 0.150 L
≈ 0.0729 M
[Acetic acid] = Initial amount of acetic acid - Moles neutralized / Volume (L)
= 0.005 moles - 0.000937 moles / 0.150 L
≈ 0.0271 M
Now, let's find the final pH using the Henderson-Hasselbalch equation.
pH_final = 4.76 + log(0.0729 / 0.0271)
Using a calculator, we get:
pH_final ≈ 5.17
The change in pH can be determined by subtracting the initial pH from the final pH:
Change in pH = pH_final - pH_initial
= 5.17 - 4.76
≈ 0.41
Therefore, the change in pH after adding 0.937 mL of 1 M sodium hydroxide to the buffer solution is approximately 0.41.
Umm... Im in something grade. I don't know this stuff.
hi