Chemistry Question

[MrCodeDude]

The pH of a 0.175M solution of the hypothetical acid HOW is 5.87. Calculate the Ka for this acid.

HOW (aq) <--> (represents equilibrium) H+(aq) * OW-(aq)

Here are steps to solving the problem.
1. Write the equilibrium equation and Ka expression for the acid (already done)
2. Convert pH to [H+] ( equation is: pH = -log [H+] )
3. Determine the concentrations of all substances in the equilibrium equation.
4. Substitute into Ka expression and solve for Ka.

I got my answer to come out to be 1.04 x 10^-11, while my friend got 9.6 x 10^-12

Thanks in advance to anyone who helps
-- mrcodedude

 

[MrCodeDude]

! Bump

 

[Siddhartha]

Originally posted by: MrCodeDude
The pH of a 0.175M solution of the hypothetical acid HOW is 5.87. Calculate the Ka for this acid.

HOW (aq) <--> (represents equilibrium) H+(aq) * OW-(aq)

Here are steps to solving the problem.
1. Write the equilibrium equation and Ka expression for the acid (already done)
2. Convert pH to [H+] ( equation is: pH = -log [H+] )
3. Determine the concentrations of all substances in the equilibrium equation.
4. Substitute into Ka expression and solve for Ka.

I got my answer to come out to be 1.04 x 10^-11, while my friend got 9.6 x 10^-12

Thanks in advance to anyone who helps
-- mrcodedude

Thanks for the bad C105 flash back.

 

[MrCodeDude]

No problem

 

[Indolent]

5.87=-log[x]
x=1.3*10^-6

ka=(x^2)/.175

ka=1.04*10^-11

yours would be the correct answer

 

[GtPrOjEcTX]

the two answers are very close, perhaps one of you just didn't hold the correct amount of sig figs?

 

[Legendary]

Originally posted by: Indolent
5.87=-log[x]
x=1.3*10^-6

ka=(x^2)/.175

ka=1.04*10^-11

yours would be the correct answer

Ding ding ding this is exactly how you do it.
MrCodeDude you are the WINNAR!

Err: and Indolent too.

 

[jst0ney]

Originally posted by: Indolent
5.87=-log[x]
x=1.3*10^-6

ka=(x^2)/.175

ka=1.04*10^-11

yours would be the correct answer