Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.
1 answer:
The molar solubility is 7.4× M and the solubility is 7.4×
g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→
+
S
- S S
Ksp = [ ] [
] = [S] [ S ] =
S = √ Ksp = √ 5. 5× = 7.4×
The solubility =7.4× g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of is dissociates and form one mole of each
and
ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4× mol/L = 7.4×
M
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HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
![K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%20H_%7B3%7D%20%20O%5E%7B%2B%7D%20%5D%5BO%20Cl%5E%7B-%7D%20%5D%7D%7BHClO%7D%20)
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
![K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Bx%20%5D%5Bx%5D%7D%7B1.0%20%5C%20x%20%5C%20%2010%5E%7B-4%7D%20-%20x%20%7D)
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-
HClO + H2O ⇄ H3O+ +OCl-
Then the equilibrium constant, Ka, of dilute HClO would be:
Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
HClO + H2O ⇄ H3O+ + OCl-
I 1.0x10^-4 0 0
C -x +x +x
E (1.0x10^-4 - x) x x
Substituting the excess (E) concentration to the Ka equation:
Simplifying the equation would yield a quadratic equation:
The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.
x = 1.688 x 10^-6 M = [H3O+] = [ClO-]
Then, you can determine the conc of [OH-] through pH.
pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M
Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M
Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.
Thus in relative amounts, the order would be
H2O >>> HClO > H3O+ = ClO- > OH-