2. Solve for pH of a solution that is 0.150 M of HClO.pH = -log[H+]
Relevant chemical equation:
HClO
H+ + ClO-
At equilibrium:
Ka = ( [H+][ClO-] ) / [HClO]
Set up an ICE table (initial, change, equilibrium)
HClO
H+ + ClO-
0.150M 0 0
- x + x + x
0.150-x x x
Ka = (x^2)/(0.150-x)
0.150*Ka - Ka*x = x^2
x^2 + Ka*x - 0.150*Ka = 0
Solving the quadratic:
a = 1
b = Ka
c = -0.150*Ka
x = ( -Ka +- sqrt(Ka^2 + 4*0.150*Ka) ) / 2
x = 6.59*10^-5 , -6.60*10^-5
Negative solution is rejected as it is impossible for a concentration value.
x = 6.59*10^-5 = [H+]
pH = -log[H+] = -log(6.59*10^-5) =
pH = 4.18
3. Solve for the pH of a solution that is 0.150 LiClO.Relevant chemical equations:
LiClO -> Li+ + ClO- [LiClO will fully dissociate to form these products]
ClO- + H2O -> HClO + OH- [This is an equilibrium]
Setting up an ice table:
ClO- + H2O -> HClO + OH-
0.150 --- 0 0
- x --- + x + x
0.150-x --- x x
Ka for HClO = 2.9*10^-8
Kw = 1*10^-14
Kb = Kw/Ka = 3.45*10^-7
Kb = [HClO][OH-]/[ClO-]
Kb = x^2/(0.150-x)
0.150*Kb - Kb*x = x^2
x^2 + Kb*x - 0.150*Kb = 0
Solving the quadratic:
a = 1
b = Kb
c = -0.150*Kb
x = ( -Kb +- sqrt(Kb^2 + 4*0.150*Kb) ) / 2
x = 2.27*10^-4, -2.28*10^-4
Negative solution is rejected as it is impossible for a concentration value.
x = 2.27*10^-4 = [OH-]
pH = 14 - pOH
pOH = -log[OH-]
pH = 14 + log(2.27*10^-4)
pH = 10.4
4. Would a solution of NH4ClO be acidic, basic, or neutral?NH4ClO will dissociate in solution to form NH4+ and ClO- ions.
The associated equilibria with these two ions are as follows:
NH4+ +H2O -> NH3 + H3O+
ClO- + H3O+ -> HClO + H2O
NH4+ will donate a proton while ClO- will accept a proton; therefore we can compare the Ka value of NH4+ to the Kb value for ClO- to determine whether the solution will be acidic, basic, or neutral.
NH4+ is the conjugate acid of NH3, therefore we can use the Kb of NH3 to find the Ka for NH4+.
Kw = 1*10^-14
Kb = 1.76*10^-5
Ka*Kb=Kw
Ka = Kw/Kb = 5.68*10^-10
ClO- is the conjugate base of HClO, therefore we can use the Ka of HClO to find the Kb for ClO-.
Kw = 1*10^-14
Ka = 2.9*10^-8
Kb = Kw/Ka = 3.45*10^-7
Now we compare the two values:
Ka = 5.68*10^-10
Kb = 3.45*10^-7
Kb is significantly larger than Ka, therefore there will be a net decrease in [H+] and the solution will be basic.
5. Solve for the pH of 1.00 L solution that has 0.150 M HClO and 0.150 M LiClO.Setting up an ice table:
HClO -> H+ + ClO-
0.150 0 0.150
- x + x + x
0.150-x x 0.150+x
Ka = (0.150+x)x/(0.150-x)
0.150*Ka - Ka*x = 0.150x + x^2
x^2 + (0.150+Ka)x - 0.150*Ka = 0
Solving the quadratic:
a = 1
b = 0.150+Ka
c = -0.150*Ka
x = ( -0.150-Ka +- sqrt((0.150+Ka)^2 + 4*0.150*Ka) ) / 2
x = 2.9*10^-8, -0.15
Negative solution is rejected as it is impossible for a concentration value.
Note: This solution can be simplified if we assume that 0.150 + Ka = 0.150 and 0.150 - Ka = 0.150 since Ka is so small.
x = [H+]
pH = -log[H+] = -log(2.9*10^-8) = 7.54
This post was edited by St4tiK on Mar 24 2010 10:44am