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Mar 24 2010 10:05am
given
HClO (ka = 2.9 * 10^-8)
NH3 (kb = 1.76 * 10^-5)

1. what is conjugate base of HCLO
=CLO- (got this i think)

2. Solve for ph of solution that is 0.150 M of HClO.
3. Solve for ph of solution that is .150 M of LiClO
4. Would a solution of NH4ClO be acidic, base, or neutral?

5. Solve for the ph of 1.00 L solution that has 0.150 M HClO and 0.150 M LiClO.
6. For the solution in question above, what would be the ph if 0.400 g of NaOH is added to the solution?



7. Using only the info in on here describe a buffer that would have:
1.000 L of solution
ph of 9.246
not exceed a pH of 9.300 when 1 g of NaOH is added
indentify the conjugate acid-base pair that would compose this buffer and the concentration of each


paying 10 fg per question excluding #1
i'd usually won't have any trouble solving these but forgot this hw was due today
and no time to look back and study lol

must show work for payment ofc


This post was edited by iReflux on Mar 24 2010 10:06am
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Mar 24 2010 10:15am
I know alll of this... im going to my class now but pm me and ill give you answer later
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Mar 24 2010 10:18am
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
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Mar 24 2010 10:19am
nh4clo = basic


thats akll i have time for lol

This post was edited by jbru5 on Mar 24 2010 10:20am
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Mar 24 2010 10:25am
Quote (St4tiK @ Mar 24 2010 12:18pm)
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 - 0.150Ka = 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


this is correct ^^
10 fg sent
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Mar 24 2010 10:43am
got 1-5 down

need 6 n 7

This post was edited by iReflux on Mar 24 2010 10:43am
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Mar 24 2010 10:50am
got #6

need 7
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Mar 24 2010 07:09pm
Quote (iReflux @ Mar 24 2010 11:25am)
this is correct ^^
10 fg sent


St4tik raped the fuck out of that question for only 10. Would help but.... you've got 10k fg and are stingy.




I'm a little baby. :baby:
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