.12M Lactic Acid (HC3H5O3, Ka=1.4x10^-4) is mixed with .10M NaC3H5O3 to form 1.00L solution.
A. Calculate the pH of the solution after the addition of 50.0mL of 1.00M NaOH.
B. Calculate the pH of the solution after the addition of 120.0mL of 1.00M NaOH.
It would be nice to know how much lactic acid and how much sodium acetate were used to make the 1 L of solution.
it doesn't say :(. that's the problem word-for-word
To calculate the pH of the solution after the addition of NaOH, we first need to determine the initial concentration of lactic acid (HC3H5O3) and its conjugate base (C3H5O3-).
A. Calculating the Initial Concentration:
From the information given, we know that we have a solution with 0.12 M lactic acid (HC3H5O3) and 0.10 M sodium lactate (NaC3H5O3).
Since lactic acid is a weak acid, we need to consider its dissociation reaction in water:
HC3H5O3 ⇌ H+ + C3H5O3-
The concentration of the conjugate base (C3H5O3-) can be calculated by using the formula:
[C3H5O3-] = [NaC3H5O3]
Therefore, the initial concentration of C3H5O3- is also 0.10 M.
B. Calculating the moles of NaOH added:
We are now adding 50.0 mL (0.050 L) of 1.00 M NaOH. Using the formula:
Molarity (M) = Moles (mol) / Volume (L)
We can calculate the moles of NaOH added:
Moles of NaOH = Molarity (M) * Volume (L)
Moles of NaOH = 1.00 M * 0.050 L
Moles of NaOH = 0.050 mol
C. Calculating the moles of HC3H5O3 and C3H5O3- remaining:
NaOH reacts with HC3H5O3 (lactic acid) to form water (H2O) and C3H5O3- (conjugate base). The balanced equation for this neutralization reaction is:
HC3H5O3 + NaOH → H2O + NaC3H5O3
From the balanced equation, we can see that 1 mol of HC3H5O3 reacts with 1 mol of NaOH.
Since the moles of HC3H5O3 and NaOH are equal, and the initial concentration of HC3H5O3 is 0.12 M, the moles of HC3H5O3 remaining after the reaction can be calculated as:
Moles of HC3H5O3 remaining = Initial moles of HC3H5O3 - Moles of NaOH added
Moles of HC3H5O3 remaining = 0.12 M * 1.00 L - 0.050 mol
Moles of HC3H5O3 remaining = 0.07 mol
Similarly, as the reaction proceeds, the moles of C3H5O3- formed can be calculated as:
Moles of C3H5O3- formed = Moles of NaOH added
Moles of C3H5O3- formed = 0.050 mol
D. Applying the Henderson-Hasselbalch equation:
The pH of the solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([C3H5O3-] / [HC3H5O3])
where pKa is the negative logarithm of the acid dissociation constant (Ka) for lactic acid, which is given as 1.4x10^-4 in the problem.
Substituting the values,
pH = -log(1.4x10^-4) + log(0.050 mol / 0.07 mol)
pH = -log(1.4x10^-4) + log(0.714)
pH = -(-3.85) + (-0.146)
pH = 3.85 - 0.146
pH ≈ 3.704
Therefore, the pH of the solution after the addition of 50.0 mL of 1.00 M NaOH is approximately 3.704.