Explore BrainMass
Share

PH, pOH, PKw

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

Calculating pH, POH, PKa, PKb.

37) PH 0.600 M solution of Na2S
Sodium sulfide is a salt of a weak acid (hydrogen sulfide) and a strong alkali (sodium hydroxide). It therefore undergoes hydrolysis in water.
Na2S = 2Na+ + S2-
S2- + 2H2O = H2S + 2OH-
Kh = [H2S][OH-]
[S2-]
[OH-] = Kh[S2-]
[H2S]
POH = -logKh + log[H2S] - log[S2-]
PKw = PH + POH
POH = 14 - PH
14 - PH = PKh + log[H2S] - log[S2-]
PH = 14 - PKh + log [S2-] - log[H2S]
S2- + 2H2O = H2S + 2OH-
T=0 1 mol 0 0
T=t (1-x) mol x mol 2x mol
Conc. c(1-x) cx 2cx

Kh = [H][OH-]
[CN-]
= [HCN](Kw/[H+])
[CN-]
= [HCN]Kw
[H+][CN-]
[H+][CN-] = Kw/Kh
[HCN]
[H+][CN-] = Ka
[HCN]
Ka = Kw
Kh
Kh = Kw
Ka
Kh = cx.2cx
c(1-x)
1-x is negligible, therefore,
Kh = 2cx2
x = (Kh/2c)1/2
= (Kw/Ka.c)1/2
= (10-14/Ka x 2x6x10-1)1/2
=(10-13/12Ka)1/2
cx = [OH-]
= 6 x10-1 x (10-13/12Ka)1/2
POH= -log [6 x10-1 x (10-13/12Ka)1/2]
= 1 - log6 + ½(13 + log 12 + log Ka)
= 1 - 0.7782 + 6.5 + 1.0792 + log Ka
= 7.801 + log Ka
PH = pKw - pOH
= (14 - 7.801) - log Ka
= PKa + 6.199
Substitute pKa for H2S.

© BrainMass Inc. brainmass.com October 24, 2018, 8:37 pm ad1c9bdddf
https://brainmass.com/chemistry/physical-chemistry/ph-poh-pkw-102472

Attachments

Solution Preview

See the attachment.

37) PH 0.600 M solution of Na2S
Sodium sulfide is a salt of a weak acid (hydrogen sulfide) and a strong alkali (sodium hydroxide). It therefore undergoes hydrolysis in water.
Na2S = 2Na+ + S2-
S2- + 2H2O = H2S + 2OH-
Kh = [H2S][OH-]
[S2-]
[OH-] = ...

Solution Summary

PH, pOH, PKw calculations are provided. Sodium sulfide as a weak acid is determined.

$2.19
See Also This Related BrainMass Solution

Systematic Treatment of Equilibrium

Solve the following problems using the systematic approach to equilibrium problems.
1. Determine the concentration of Ag+, CN-, and HCN in a saturated solution of AgCN where the pH is fixed at 5.90.
2. Determine the concentration of NH4+, NH3, CH3COOH (acetic acid), and CH3COO- (acetate ion) if 0.0050 moles of ammonium acetate (NH4+ -OOCCH3) is dissolved in 1.00 L of water and the pH is fixed at 5.90.
3. Calculate the pH of 5.0 x 10-8 M HCl. You will need to use the systematic treatment of equilibrium, and there are no simplifying assumptions you can make. Approximately what fraction of the total H+ in this solution is due to the autoprotolysis of water?
4. Write chemical reactions for the following:
a. Ka for 3-nitrobenzoic acid
b. Kb for 3-nitrobenzoate
c. Kb for benzylamine
d. Ka for benzylammonium ion
5. Calculate the pH and fraction of dissociation for a 0.10 M solution of formic acid.
6. Calculate the pH and fraction of association for a 0.10 M solution of sodium formate.
7. Calculate the pH and fraction of dissociation for a 0.050 M solution of ethanolamine (2-amino ethanol).

View Full Posting Details