a 275mL sample of vapor in equilibrium with 1-propylamine at 25C is removed and dissolved in 0.500L of H20. For 1-propalamine, Kb=3.72*10^-4 and the vapor pressure at 25C is 215 torr.
(a)What should be the pH of the aqueous solution?
(b) How many mg of NaOH dissolved in 0.500L of water give the same pH?
Could you please give the exact steps to go through this problem? (answer too, if you can)thanks
I worked this for someone just a second ago. If not for you, repost at the top of the board and I'll find it to give you a link.
http://www.jiskha.com/display.cgi?id=1300829574
To solve this problem, we need to follow these steps:
Step 1: Calculate the moles of 1-propylamine in the vapor phase
We know that the sample is in equilibrium with 1-propylamine. To calculate the moles, we need to use the ideal gas law: PV = nRT.
Rearranging the equation, we have n = PV/RT, where n is the number of moles, P is the pressure (in atm), V is the volume (in L), R is the ideal gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin (25C + 273 = 298K).
Given:
Pressure (P) = 215 torr = 215/760 atm
Volume (V) = 275 mL = 0.275 L
Temperature (T) = 25C + 273 = 298K
Calculating n:
n = (215/760) * 0.275 / 0.0821 * 298
n ≈ 0.00392 moles
Step 2: Calculate the concentration of 1-propylamine in the aqueous solution
To find the concentration, we use the following formula:
Kb = [OH-][1-propylamine] / [H2O]
Since the concentration of water remains constant, we can ignore it and rewrite the equation as:
Kb = [OH-][1-propylamine]
Given:
Kb = 3.72 * 10^-4
Let's assume [1-propylamine] = x (M)
Since OH- is produced in a 1:1 ratio, its concentration will also be x (M).
Plugging in the values into the equation:
3.72 * 10^-4 = x * x
3.72 * 10^-4 = x^2
Taking the square root of both sides:
x ≈ 0.0193 M
Step 3: Calculate the pOH
pOH = -log10([OH-])
pOH = -log10(0.0193)
pOH ≈ 1.71
Step 4: Calculate the pH
Since pH + pOH = 14, we can find the pH using the pOH value we calculated in the previous step:
pH = 14 - pOH
pH ≈ 14 - 1.71
pH ≈ 12.29
So, the answer to (a) is: The pH of the aqueous solution is approximately 12.29.
Step 5: Calculate the amount of NaOH (in mg) needed to give the same pH
To find the amount of NaOH, we first need to calculate the concentration of OH-. Since we have already calculated it as 0.0193 M, we can use this concentration to find the amount of NaOH.
Given:
Volume (V) = 0.500 L
Concentration (C) = 0.0193 M
We can use the formula:
Amount of substance (n) = Concentration (C) * Volume (V)
n = 0.0193 * 0.500
n ≈ 0.00965 moles
Now, we need to convert moles to milligrams (mg):
molecular weight of NaOH = 23 + 16 + 1 = 40 g/mol
Amount in grams = moles * molecular weight
Amount in grams = 0.00965 * 40
Amount in grams ≈ 0.386 g
And finally, we convert grams to milligrams by multiplying by 1000:
Amount in milligrams = 0.386 * 1000
Amount in milligrams ≈ 386 mg
So, the answer to (b) is: Approximately 386 mg of NaOH dissolved in 0.500 L of water will give the same pH.