A 500 ml buffer solution contains .2M Acetic acid and .3M sodium acetate. Find the pH of the buffer solution after adding 20 ml of 1M NaOH, what is the pH?
(pka = 4.74)
a) pH = pKa + [(base)/(acid)] will get the pH at the beginning.
b)HAc + OH^- ==> H2O + Ac^-
initial:
HAc = 500 x 0.2 M = 100 millimoles.
OH^- = 0
Ac^- = 500 x 0.3M = 150 millimoles.
change:
OH^- = 20 mL x 1 M = 20 millimoles added
HAc = 100-20
Ac^- = 150 + 20
final:
HAc = 80 millimoles
Ac^- = 170 millimiles
Convert millimoles to concn, substitute into the HH equation above and solve for pH.
To find the pH of the buffer solution after adding NaOH, we need to calculate the change in concentrations of acetic acid (CH3COOH) and acetate ion (CH3COO-) due to the reaction with NaOH.
First, let's calculate the initial concentrations of acetic acid and acetate ion in the buffer solution:
Volume of acetic acid (V1) = 500 ml
Volume of sodium acetate (V2) = 500 ml
Total volume of the buffer solution = V1 + V2 = 1000 ml = 1 liter
Concentration of acetic acid (C1) = 0.2 M
Concentration of sodium acetate (C2) = 0.3 M
Now, we add 20 ml of 1 M NaOH to the buffer solution.
To calculate the final concentrations of acetic acid and acetate ion, we need to consider the stoichiometry of the reaction between acetic acid and NaOH:
CH3COOH + NaOH → CH3COONa + H2O
Since the stoichiometry is 1:1, 1 mole of acetic acid reacts completely with 1 mole of NaOH.
Initial moles of acetic acid (moles1) = C1 * V1 = 0.2 M * 0.5 L = 0.1 moles
Moles of NaOH added (moles2) = C3 * V3 = 1 M * 0.02 L = 0.02 moles
Due to the reaction, 0.02 moles of acetic acid will be neutralized, and an equivalent amount of acetate ions will be formed.
Final moles of acetic acid (moles1-final) = moles1 - moles2 = 0.1 moles - 0.02 moles = 0.08 moles
Final moles of acetate ion (moles2-final) = moles2 = 0.02 moles
Now, let's calculate the final concentrations of acetic acid and acetate ion:
Final concentration of acetic acid (C1-final) = moles1-final / V1 = 0.08 moles / 0.5 L = 0.16 M
Final concentration of acetate ion (C2-final) = moles2-final / V2 = 0.02 moles / 0.5 L = 0.04 M
To calculate the pH of the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log10(C2-final / C1-final)
pKa = 4.74 (given)
pH = 4.74 + log10(0.04 M / 0.16 M)
pH = 4.74 + log10(0.25)
pH ≈ 4.74 + (-0.602)
pH ≈ 4.14
Therefore, the pH of the buffer solution after adding 20 ml of 1M NaOH is approximately 4.14.
To find the pH of the buffer solution after adding NaOH, we need to consider the reaction that occurs between the added NaOH and the components of the buffer solution.
The reaction is as follows:
CH3COOH (acetic acid) + OH- (from NaOH) -> CH3COO- (acetate ion) + H2O
Initially, before adding NaOH, we have a mixture of acetic acid and sodium acetate. The acetic acid acts as a weak acid, and sodium acetate acts as its conjugate base (weak base).
In a buffer solution, the pH is determined by the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH = the pH of the buffer solution
pKa = the acid dissociation constant of acetic acid (given as 4.74)
[A-] = concentration of the conjugate base (acetate ion)
[HA] = concentration of the weak acid (acetic acid)
Given that the initial concentrations of acetic acid and sodium acetate are 0.2M and 0.3M, respectively, we can calculate the values of [A-] and [HA].
[A-] = [sodium acetate] = 0.3M
[HA] = [acetic acid] = 0.2M
Now, let's consider the effect of adding 20 ml (0.02L) of 1M NaOH to the buffer solution. In this case, we need to calculate how much of the acetic acid and sodium acetate will react and convert into their conjugate base and acid, respectively.
From the stoichiometry of the reaction, we can determine that for every mole of NaOH added, one mole of acetic acid is converted to acetate ion. So, the amount of acetic acid that will react can be calculated as follows:
moles of acetic acid = concentration of acetic acid × volume of acetic acid (before addition of NaOH)
moles of acetic acid = 0.2M × 0.5L (initial volume of buffer solution)
Similarly, one mole of acetic acid converted results in one mole of acetate ion formed. So, the amount of acetate ion formed can be calculated as:
moles of acetate ion = moles of acetic acid reacted with NaOH
Now, we need to calculate the concentrations of [A-] and [HA] in the final buffer solution.
[A-] = [sodium acetate] + moles of acetate ion formed/molal volume of buffer solution (initial volume + volume of NaOH added)
[HA] = [acetic acid] - moles of acetic acid reacted with NaOH/molal volume of buffer solution (initial volume + volume of NaOH added)
Substituting the initial concentrations and calculated values of [A-] and [HA] into the Henderson-Hasselbalch equation, we can find the pH of the buffer solution after adding NaOH.