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Write a balanced chemical equation for the dissociation of hydrazoic acid, HN3, in water. What are the pH and percent dissociation of a 2.0 molar solution of hydrazoic acid that contains 1.50 g of calcium azide (Ca(N3)2) in 100 mL of solution. Ka hydrazoic acid= 3.6 X 10^-6 COMMON ION EFFECT

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Related questions

A 0.200M solution of hydrazoic acid (HN3) is prepared. The Ka for HN3 is 2.57x10^-5 and the Kb for the azide ion (N3) is 3.98x10^-10.

a) Write the dissociation equation for HN3 and identify the conjugate acid/base pairs.

b) Calculate the pH of the solution.

c) Calculate the Percent Ionization of the HN3 solution.

Please show work and explain all parts i need to know this for my final!

Ignoring activities, determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 3.696. Ksp (CuN3) = 4.9x10-9 ; Ka (HN3)= 2.2x10-5.

List of steps my teacher suggests we follow:

1. Write the pertinent reactions.

2. Write the charge balance equation.

3. Write the mass balance equations.

4. Write the equilibrium constant expressions for each reaction.

5. Count the equations and unknowns.

6. Solve for all unknowns.

How far I've gotten:

1. Pertinent reactions:

CuN3 (s) <-> Cu+ (aq) + N3- (aq) Ksp= 4.9x10-9

N3- (aq) + H2O (l) <-> HN3 (aq) + OH- (aq) Kb = (Kw/Ka) = (1.0x10-14/2.2x10-5) = 4.55 x 10-10

H2O (l) <-> H+ (aq) + OH- (aq) Kw = 1.0x10-14

2. Write the charge balance equation:

Because the pH is fixed with no info given regarding the buffer used to fix the pH, there is no charge balance equation for this problem.

3. Write the mass balance equation(s):

[Total Cu+] = [Total N3-]

[Cu+] = [N3-] +[HN3]

4. Write the equilibrium constant expressions for each reaction:

Ksp = 4.9x10-9 = [Cu+] [N3-]

Kb = 4.55x10-10 = ([HN3] [OH-] / [N3-])

Kw = 1.0x10-14 = [H+] [OH-]

Now use the charge balance, mass balance, and equilibrium constant expressions to solve for [Cu+], which is the molar solubility of CuN3.

I tried to finish it 5 times and haven't been able to make any progress.

What is the pH of a solution made by dissolving 4.3 g of K2CO3 and 1.6 g of KHCO3 in 500. mL of water? (See the appendix.)

Table 3 An Acid-base table
Acid Name ConjugateAcid Ka pKa Conjugate Base Base Name
perchloric acid HClO4 >>1 < 0 ClO41- perchlorate ion
hydrohalic acid HX (X=I,Br,Cl) >>1 < 0 X1- halide ion
sulfuric acid H2SO4 >>1 < 0 HSO41- hydrogen sulfate ion
nitric acid HNO3 >>1 < 0 NO31- nitrate ion
hydronium ion H3O1+ 1.0 0.00 H2O water
iodic acid HIO3 0.17 0.77 IO31- iodate ion
oxalic acid H2C2O4 5.9 x 10-2 1.23 HC2O41- hydrogen oxalate ion
sulfurous acid H2SO3 1.5 x 10-2 1.82 HSO31- hydrogen sulfite ion
hydrogen sulfate ion HSO41- 1.2 x 10-2 1.92 SO42- sulfate ion
phosphoric acid H3PO4 7.5 x 10-3 2.12 H2PO41- dihydrogen phosphate ion
hydrofluoric acid HF 7.2 x 10-4 3.14 F1- fluoride ion
nitrous acid HNO2 4.0 x 10-4 3.40 NO21- nitrite ion
lactic acid HC3H5O3 6.4 x 10-5 3.85 C3H5O31- lactate ion
formic acid HCHO2 1.8 x 10-4 3.74 CHO21- formate ion
hydrogen oxalate ion HC2O41- 6.4 x 10-5 4.19 C2O42- oxalate ion
hydrazoic acid HN3 1.9 x 10-5 4.72 N31- azide ion
acetic acid HC2H3O2 1.8 x 10-5 4.74 C2H3O21- acetate ion
carbonic acid H2CO3 4.3 x 10-7 6.37 HCO31- hydrogen carbonate ion
hydrogen sulfite ion HSO31- 1.0 x 10-7 7.00 SO32- sulfite ion
hydrosulfuric acid H2S 1.0 x 10-7 7.00 HS1- hydrogen sulfide ion
dihydrogen phosphate ion H2PO41- 6.2 x 10-8 7.21 HPO42- hydrogen phosphate ion
hypochlorous acid HClO 3.5 x 10-8 7.46 ClO1- hypochlorite ion
ammonium ion NH41+ 5.6 x 10-10 9.25 NH3 ammonia
hydrocyanic acid HCN 4.0 x 10-10 9.40 CN1- cyanide ion
hydrogen carbonate ion HCO31- 4.7 x 10-11 10.33 CO32- carbonate ion
hydrogen phosphate ion HPO42- 4.8 x 10-13 12.32 PO43- phosphate ion
hydrogen sulfide ion HS1- 1.3 x 10-13 12.89 S2- sulfide ion
water H2O 1.0 x 10-14 14.00 OH1- hydroxide ion
ammonia NH3 <<10-14 NH21- amide ion
hydroxide ion OH1- <<10-14 O2- oxide ion

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