Question

Question 1

Hydroflouric acid, HF, is a weak acid with one acidic proton. What is the pH of a 0.20 M solution of this acid? (Ka = 6.8 x 10-4 for HF.)

Question 2

Under which set of conditions does
NH3 (g) best follow the ideal gas law?

Answer 1

Question 1.

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.2 M; then

x^2 + (7.2*10^-4)x - 0.2*(7.2*10^-4) = 0

solve for x

x =7.17*10^-4

substitute

[H+] = 0 + 7.17*10^-4= 7.17*10^-4M

[A-] = 0 + 7.17*10^-4= 7.17*10^-4M

[HA] = M - x = 0.2-7.17*10^-4= 0.199283 M

pH = -log(H+) = -log(7.17*10^-4) = 3.144

Question 2.

Now...

NH3(g) is a small molecule, yet, will experience dipole/dipole interactions due to th elone pair present.

For this to avoid deviations from ideal gas law

ensure that:

moelcules are far away, which can be obtained via:

- Use P < 10 atm

- Use T > 100 K

This will make particles far away, therefore, can't interact

Which means this will follow

PV = nRT