A 275 mL sample of 1-propylamine vapor (in equilibrium with the liquid) at 25 °C is removed
and dissolved in 0.500 L of H2O. For 1-propylamine, Kb = 3.72 x 10-4
and the vapor pressure at
25.0 °C is 316 Torr.
a. What should be the pH of the aqueous solution?
b. How many milligrams of NaOH dissolved in 0.500 L of water give the same pH?
1. Use ideal gas law to determine moles of ProAm.
Be sure to convert given data to units of R-Value.
2. Determine concentration in 0.500L soln. =[moles ProAm/Volume(L)].
3. Determine [OH]=SqrRt([ProAm]Kb)
4. Determine pOH=-log[OH]
5. Determine pH=14-pOH
(I got pH = 11.5).
PART II:
1. [NaOH] => pH=11.5 => Same[OH]in Part I.
2. Determine moles NaOH in 0.500L = 1/2[OH] fm part I.
3. Multiply by 40 gms/mole then multiply by 1000 => mg. NaOH.
(I got 66mg/0.50L)
To find the pH of the aqueous solution in part a, we need to calculate the concentration of hydroxide ions (OH-) in the solution. Given that the vapor pressure of 1-propylamine at 25 °C is 316 Torr and the Kb value is 3.72 x 10^-4, we can use the equilibrium constant expression for the reaction of 1-propylamine with water:
1-propylamine + H2O ⇌ 1-propylammonium ion + OH-
Step 1: Calculate the concentration of 1-propylamine in moles per liter (M).
First, let's convert the vapor pressure of 1-propylamine to atm by dividing it by 760 Torr:
316 Torr / 760 Torr/atm = 0.416 atm
Next, we can use the ideal gas law to calculate the number of moles of 1-propylamine in the vapor phase:
PV = nRT
Using the values for pressure (P), volume (V), gas constant (R), and temperature (T):
0.416 atm * 0.275 L = n * 0.0821 L·atm/(mol·K) * 298 K
Solving for n (number of moles):
n = (0.416 atm * 0.275 L) / (0.0821 L·atm/(mol·K) * 298 K) = 0.0129 mol
The concentration of 1-propylamine is:
0.0129 mol / 0.500 L = 0.0258 M
Step 2: Calculate the concentration of OH- ions in the solution.
Using the Kb value and the concentration of 1-propylamine, we can write an equation to find the concentration of OH-:
Kb = [1-propylammonium ion][OH-] / [1-propylamine]
3.72 x 10^-4 = x * x / (0.0258 - x)
Assuming x << 0.0258:
3.72 x 10^-4 = x^2 / 0.0258
Solving for x (concentration of OH-):
x = sqrt((3.72 x 10^-4) * 0.0258) = 1.46 x 10^-3 M
Step 3: Calculate the pH of the solution.
The concentration of OH- is equal to the concentration of the hydroxide ions in the solution. We can use the relationship between pH and pOH to find the pH:
pOH = -log[OH-]
pOH = -log(1.46 x 10^-3) ≈ 2.84
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH = 14 - 2.84 ≈ 11.16
Therefore, the pH of the aqueous solution is approximately 11.16 (part a).
For part b, we need to calculate how many milligrams of NaOH will give the same pH as the previous solution. NaOH completely dissociates in water, giving one hydroxide ion for each NaOH molecule.
Step 1: Calculate the moles of OH- ions needed.
From part a, we found that the concentration of OH- ions is 1.46 x 10^-3 M. Since the volume of the solution is 0.500 L, the moles of OH- ions needed can be calculated as:
moles of OH- = concentration of OH- ions * volume of solution
= 1.46 x 10^-3 mol/L * 0.500 L
= 7.3 x 10^-4 mol
Step 2: Convert moles to milligrams.
Since the molar mass of NaOH is 22.99 + 16.00 + 1.00 = 39.99 g/mol, we can calculate the mass of NaOH needed as:
mass of NaOH = moles of OH- ions * molar mass of NaOH
= 7.3 x 10^-4 mol * 39.99 g/mol
= 2.92 x 10^-2 g
Converting grams to milligrams:
mass of NaOH = 2.92 x 10^-2 g * 1000 mg/g
= 29.2 mg
Therefore, approximately 29.2 milligrams of NaOH dissolved in 0.500 L of water will give the same pH as the previous solution (part b).