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
8 answers
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.
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
[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.
2 answers