reposted again

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

how do i now which values to use for part b? for a i got pH of 11.2

its 316 not 215 torr

pH = 11.2. Convert to pOH.

pH + pOH = pKw = 14, then convert pOH to (OH^-) and that will be in moles/L.
For 0.5 L you will have half that number of moles. grams = moles x molar mass
Then convert grams to mg.

will it be 3.17*10^-3mg?

I find those digits but I have a different position for the decimal. If I didn't goof it is 31.7 mg.

so the correct answer is 31.7mg? I don't see what i did wrong. Could You show me the steps please?

If you will post you work I will look for the error.

pH=11.23

[H+]= 10^(-11.23)=5.888*10^-12
(1*10^-14)/(5.888*10^-12)=[OH]=.001698M/L
.001698/2= .0008491 moles/.5L

.0008491*40(molar mass of NaOH)=.033964*1000= 33.96mg

I agree with 33.96 when using 11.23 for pH but I would round it to 34.0 mg to three s.f.. My answer of 31.7 mg is correct using 11.2 for part a.

To determine the pH of the aqueous solution in part (a), you need to calculate the concentration of hydroxide ions (OH-) in the solution. Since the initial sample was in equilibrium with the vapor phase, it means that the concentration of 1-propylamine will also be equal to the concentration of OH- ions in the aqueous solution.

First, let's find the concentration of 1-propylamine in the vapor phase. We can use the ideal gas law to do this:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles of gas
R = ideal gas constant
T = temperature (in Kelvin)

We know that the volume (V) is 275 mL = 0.275 L and the pressure (P) is 215 torr. The temperature (T) is given as 25°C, which we need to convert to Kelvin.

T(K) = T(°C) + 273.15
T(K) = 25 + 273.15 = 298.15 K

Now, we rearrange the ideal gas law equation to solve for n:

n = PV / RT

Calculating the value of n, we have:

n = (215 torr * 0.275 L) / (0.0821 atm L/mol K * 298.15 K)
n = 0.0277 mol

Since the initial sample was dissolved in 0.500 L of water, the concentration of 1-propylamine (and OH- ions) in the aqueous solution will be:

[OH-] = n / V
[OH-] = 0.0277 mol / 0.500 L
[OH-] = 0.0554 M

Now, to find the pH, we need to calculate the pOH using the pOH equation:

pOH = -log [OH-]

pOH = -log (0.0554) = 1.26

Since pH + pOH = 14, we can find the pH:

pH = 14 - pOH
pH = 14 - 1.26
pH = 12.74

So, the pH of the aqueous solution is approximately 12.74.

Moving on to part (b), to find how many mg of NaOH dissolved in 0.500 L of water will give the same pH, we need to find the concentration of OH- ions required to achieve that pH.

Using the equation:

pOH = -log [OH-]

Rearranging for [OH-]:

[OH-] = 10^(-pOH)

[OH-] = 10^(-12.74) = 4.22 × 10^(-13) M

Now, let's calculate the number of moles of OH- ions:

mol OH- = [OH-] * volume (in L)
mol OH- = 4.22 × 10^(-13) M * 0.500 L
mol OH- = 2.11 × 10^(-13) mol

Since we want to find the mass of NaOH required, we need to consider the molar mass of NaOH:

Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Now, we can calculate the mass (in grams) of NaOH required:

mass NaOH = mol OH- * molar mass NaOH
mass NaOH = 2.11 × 10^(-13) mol * 40.00 g/mol
mass NaOH = 8.44 × 10^(-12) g

Finally, to convert the mass to milligrams:

mass NaOH = 8.44 × 10^(-12) g * (1 mg / 0.001 g)
mass NaOH ≈ 8.44 × 10^(-9) mg

Therefore, approximately 8.44 × 10^(-9) mg of NaOH dissolved in 0.500 L of water will give the same pH as calculated in part (a).