(a)What is the pH of a solution made by combining 200 mL of 0.1 M Lactic Acid with 300 mL of 0.075 M sodium lactate? (b) If 20 mL of 0.4 M HCl is added to the solution from (a) what is the resulting pH?
Use the Henderson-Hasselbalch equation for a.
pH = pKa + log(base)/(acid)
b.
millimols acid = 200 x 0.1 = 20
millimols base = 300 x 0.075 = 22.5
millimols HCl added = 20 x 0.4 = 8
Set up an ICE chart.
.........lactate + H^+ ==> lactic acid
I..........22.5....0........20
added..............8...........
C..........-8.....-8........+8.....
E..........14.5....0........28
Plug the E line into another HH equation and solve for pH. Adding acid should decrease the pH.
To determine the pH of a solution, we need to calculate the concentration of hydrogen ions (H+) in the solution. pH is defined as the negative logarithm (base 10) of the concentration of H+ ions.
Let's start by calculating the concentration of H+ ions in the solution in part (a).
(a) To find the concentration of H+ ions in the solution, we need to consider the dissociation of lactic acid (HA) and sodium lactate (A-).
The dissociation of lactic acid (HA) is as follows:
HA ⇌ H+ + A-
The dissociation of sodium lactate (A-) is as follows:
A- ⇌ Na+ + OH-
Since lactic acid (HA) is a weak acid, it does not completely dissociate, so we can use the Henderson-Hasselbalch equation to calculate the concentration of H+:
pH = pKa + log([A-]/[HA])
For lactic acid, pKa = 3.86, which represents the acidity constant. The ratio [A-]/[HA] can be calculated using the amount of each substance and their respective concentrations.
Given:
- Volume of lactic acid (HA) = 200 mL
- Concentration of lactic acid (HA) = 0.1 M
- Volume of sodium lactate (A-) = 300 mL
- Concentration of sodium lactate (A-) = 0.075 M
First, we need to convert the volumes to liters:
- Volume of lactic acid (HA) = 200 mL = 0.2 L
- Volume of sodium lactate (A-) = 300 mL = 0.3 L
Next, we calculate the moles of each substance:
- Moles of lactic acid (HA) = Volume (L) × Concentration (M) = 0.2 L × 0.1 M = 0.02 moles
- Moles of sodium lactate (A-) = Volume (L) × Concentration (M) = 0.3 L × 0.075 M = 0.0225 moles
Now we can calculate the ratio [A-]/[HA]:
[A-]/[HA] = Moles of sodium lactate (A-) / Moles of lactic acid (HA) = 0.0225 moles / 0.02 moles = 1.125
Substituting the values into the Henderson-Hasselbalch equation:
pH = 3.86 + log(1.125)
To find the resulting pH, calculate the natural logarithm of 1.125 and multiply it by the constant 2.303:
pH = 3.86 + 2.303 × log(1.125)
Using a scientific calculator, compute the logarithm and add it to 3.86 to find the pH value.
(b) If 20 mL of 0.4 M HCl is added to the solution from part (a), the pH will change due to the change in the concentration of H+ ions in the solution.
Given:
- Volume of HCl = 20 mL = 0.02 L
- Concentration of HCl = 0.4 M
To calculate the new concentration of H+ ions in the solution, we need to consider the new amount of H+ ions added by the HCl.
Moles of HCl = Volume (L) × Concentration (M) = 0.02 L × 0.4 M = 0.008 moles
Since HCl is a strong acid, it will fully dissociate, meaning the concentration of H+ ions equals the concentration of HCl. Thus, the new concentration of H+ ions in the solution can be given as:
New concentration of H+ ions = ([H+ ions from lactic acid] + [H+ ions from HCl])
Substituting the values, we obtain:
New concentration of H+ ions = ([H+ ions from lactic acid] + 0.008 moles)
To find the new pH, we need to calculate the concentration of H+ ions based on the new moles and volume.
Finally, we can find the resulting pH by taking the negative logarithm (base 10) of the new calculated concentration of H+ ions.