ph and pKb question

[sharpieLIFE]

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All set thanks

 

[LoLCareerGoals]

NaHCO3 -> [Na+] + [HCO3-]

[HCO3-] <-> [H+] + [HCO3 2-]
Carbonic acid has Ka1 = 4.3E-7 and Ka2 = 5.6E-11. Then:
[HCO3-] <-> [H+] + [HCO3 2-], Keq = 5.6E-11.
Moles([HCO3-]) = moles(NaHCO3-) = 1.67/(23+1+12+48) = 2 moles
Molarity([HCO3-]) = 2/(0.2) = 0.1 M
using [HCO3-] <-> [H+] + [HCO3 2-], Keq = 5.6E-11, we have:
x^2/0.1 = 5.6E-11
x^2 = 5.6E-10
x = 2.36e-5
ph = -log10(2.36e-5) = 4.62

Edit, actually instead of taking the root we can do this:
-log10(x^2) = ph*2
-log10(5.6E-10) = ph*2. Here it is know that log(2) = 0.3 and log(4) = 0.6 and log(8) = 0.9, so log(5.6) is about 0.7
-(0.7 - 10) = ph*2
9.3 = 2*ph
ph = 4.65




Was this really an MCAT question? I'd like to see answer choices to see if there is a shortcut to do this.

 

[Gauss44]

NaHCO3 -> [Na+] + [HCO3-]

[HCO3-] <-> [H+] + [HCO3 2-]
Carbonic acid has Ka1 = 4.3E-7 and Ka2 = 5.6E-11. Then:
[HCO3-] <-> [H+] + [HCO3 2-], Keq = 5.6E-11.
Moles([HCO3-]) = moles(NaHCO3-) = 1.67/(23+1+12+48) = 2 moles
Molarity([HCO3-]) = 2/(0.2) = 0.1 M
using [HCO3-] <-> [H+] + [HCO3 2-], Keq = 5.6E-11, we have:
x^2/0.1 = 5.6E-11
x^2 = 5.6E-10
x = 2.36e-5
ph = -log10(2.36e-5) = 4.62

Edit, actually instead of taking the root we can do this:
-log10(x^2) = ph*2
-log10(5.6E-10) = ph*2. Here it is know that log(2) = 0.3 and log(4) = 0.6 and log(8) = 0.9, so log(5.6) is about 0.7
-(0.7 - 10) = ph*2
9.3 = 2*ph
ph = 4.65




Was this really an MCAT question? I'd like to see answer choices to see if there is a shortcut to do this.

Line 6: 1.67/(23+1+12+48)=.02 moles est. I think that's what you meant since it's corrected in the next line (.02/.2)=.1

 

[LoLCareerGoals]

I think solution is wrong regardless. Similar problem is solved above and everyone takes a different approach treating HCO3- exclusively as a base.