A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent.
100 ml of a 0.1M solution of the above acid is titrated with 0.1 M NaOH. After an amount of NaOH is added the pH= 5.2. How much NaOH was added?
( c ) The titration is continued to equivalence. What is the pH at equivalence?
I'll get you started by helping with the first part.
delta T = Kf*m
0.1930 = 1.86*m
m = 0.1930/1.86 = 0.10376 (Since the delta T is to 4 significant figures, I've use more places in my answer. I don't know if that is 0.1M or 0.100 M so watch the number of s.f. when you do this yourself.)
Since the solution is 0.1, that means the solution actually contains more particles than 1 (it is ionized slightly).
0.10376-0.1 = 0.0376 extra particles
%ion = (0.00376/0.1)*100 = about 4%
.............HA ==> H^+ + A^-
initial..0.100..0.00376..0.00376
change.....-x........x......x
equil....0.1-0.00376
Ka = set up Ka expression and substitute values from the ICE chart. Solve for Ka.
To find the Ka (acid dissociation constant) of the acid:
1. Start with the equation for the freezing point depression of a solution:
ΔT = Kf * m
Where:
ΔT is the change in freezing point
Kf is the cryoscopic constant for water (-1.86 °C/molal)
m is the molality of the solution
2. Plug in the given values:
ΔT = 0.193 °C
m = 0.1 molal
Kf = -1.86 °C/molal
3. Rearrange the equation to solve for the concentration of the solute (in this case, the acid):
ΔT = Kf * (n / m_solvent)
Where:
n is the number of moles of solute
m_solvent is the mass of the solvent
We know that 0.1 molal is equivalent to 0.1 molar, so the concentration of the solute is 0.1 M.
4. Calculate the number of moles of solute by multiplying the molality by the mass of the solvent. Assume a 100 g water sample:
n = 0.1 molal * 0.1 kg = 0.01 moles
5. Plug in the values into the rearranged equation:
0.193 °C = -1.86 °C/molal * (0.01 moles / m_water)
6. Solve for m_water:
m_water = (0.01 moles / (-1.86 °C/molal)) / 0.193 °C = 0.0538 kg
7. Convert the mass of water to volume (assuming the density of water is 1 g/mL):
V_water = m_water / ρ_water = 0.0538 L
8. Now, you have the number of moles of the weak acid and the volume of water used, which allows you to calculate the concentration of the acid:
C_acid = n_acid / V_water = 0.01 moles / 0.0538 L = 0.186 M
9. Finally, use the concentration of the weak acid to calculate the Ka value using the equation for a weak acid dissociation:
Ka = [H+]^2 / [HA]
Given that the pH of the solution after titration is 5.2, [H+] can be calculated as follows:
[H+] = 10^(-pH) = 10^(-5.2) = 6.31 × 10^(-6) M
Now, plug in the values into the Ka equation:
Ka = (6.31 × 10^(-6) M)^2 / 0.186 M
Solving this equation will give you the value of Ka.
To determine how much NaOH was added during titration:
1. Given that the pH of the solution after titration is 5.2, we can calculate the concentration of OH- ions in the solution using the equation:
pOH = 14 - pH = 14 - 5.2 = 8.8
[OH-] = 10^(-pOH) = 10^(-8.8) M
2. Since NaOH is a strong base, it fully dissociates in water. Therefore, the concentration of NaOH is equal to the concentration of OH- ions:
[NaOH] = [OH-] = 10^(-8.8) M
3. Use the volume of the acid solution (100 mL = 0.1 L) and the concentration of NaOH to calculate the amount of NaOH added:
Amount of NaOH = [NaOH] * Volume of acid solution = 10^(-8.8) M * 0.1 L
To find the pH at equivalence point:
At the equivalence point, the number of moles of the acid will be equal to the number of moles of NaOH. If we assume the volume of NaOH added to reach the equivalence point is V_eq, we can calculate the concentration of the acid at equivalence point as follows:
C_acid_eq = n_acid / V_eq = C_NaOH * V_NaOH / V_eq
Since the reaction is a 1:1 molar ratio between the acid and base, and the concentration of NaOH is 0.1 M, we can solve for V_eq:
0.1 M * V_NaOH = 0.1 M * 0.1 L = 0.01 moles
V_eq = 0.01 L
Now, we can calculate the concentration of the acid at equivalence:
C_acid_eq = 0.1 M * 0.1 L / 0.01 L = 1 M
Since the titration is at the equivalence point, the pH should be neutral, which is around 7.