Chem Weak Acid Strong Base Help

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Homework Statement


800ml sample of 2.00 molar formic acid HCOOH (or HA) ismixed with 200ml of a 4.80 molar NaOH. Calculate the [H3O+] of the resulting solution.


Homework Equations


(for buffers) pH=pKa-log(HA/A-)
Ka for formic acid is 1.8x10-4
pH=14-pOH
[H+]=(1.00x10-14)/([OH-)

The Attempt at a Solution


Well, I used an ICE chart of formic acid with water to find the concentration of hydronium ions at equilibrium, which came out to be 1.7x10-2M, and used some stoichiometry to figure out that there are 0.960 moles of NaOH, and thus, .960 moles of OH-.I subtracted the amount of hydronium ions from the moles of hydroxide ions (since the total volume happens to be one liter), and found this to be my concentration of hydroxide ions, .943M. Using the formula above to calculate hydronium ion concentration from hydroxide ions and the ionization constant for water at 25 degrees celcius, which we assume this is at, I derived the concentration of hydronium ions to be 1.1x10-14, which gives a pH of 13.97.
I was expecting a basic solution since there was so much hydroxide ions added, but this number just seems surreal. I appreciate any comments.


ICE Chart of Formic Acid:
HA + H2O <--> A- + H3O+
I 1.6 0 0
C -x +x +x
E 1.6-x x x
(sorry it all got squished together, but to clear things up, the water column was left blank)

1.8x10-4=(x2)/(1.6-x)
Assuming the x is negligabile to 1.6 gives us:
x=1.7x10-2
 

Answers and Replies

  • #2
GCT
Science Advisor
Homework Helper
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Homework Statement


800ml sample of 2.00 molar formic acid HCOOH (or HA) ismixed with 200ml of a 4.80 molar NaOH. Calculate the [H3O+] of the resulting solution.


Homework Equations


(for buffers) pH=pKa-log(HA/A-)
Ka for formic acid is 1.8x10-4
pH=14-pOH
[H+]=(1.00x10-14)/([OH-)

The Attempt at a Solution


Well, I used an ICE chart of formic acid with water to find the concentration of hydronium ions at equilibrium, which came out to be 1.7x10-2M, and used some stoichiometry to figure out that there are 0.960 moles of NaOH, and thus, .960 moles of OH-.I subtracted the amount of hydronium ions from the moles of hydroxide ions (since the total volume happens to be one liter), and found this to be my concentration of hydroxide ions, .943M. Using the formula above to calculate hydronium ion concentration from hydroxide ions and the ionization constant for water at 25 degrees celcius, which we assume this is at, I derived the concentration of hydronium ions to be 1.1x10-14, which gives a pH of 13.97.
I was expecting a basic solution since there was so much hydroxide ions added, but this number just seems surreal. I appreciate any comments.


ICE Chart of Formic Acid:
HA + H2O <--> A- + H3O+
I 1.6 0 0
C -x +x +x
E 1.6-x x x
(sorry it all got squished together, but to clear things up, the water column was left blank)

1.8x10-4=(x2)/(1.6-x)
Assuming the x is negligabile to 1.6 gives us:
x=1.7x10-2

Formic acid is neutralized by the hydroxide its mole value is

1.6 - 0.96

Divide this by the new volume to find the new concentration of formic acid. Then use the Ka equation as you have set up to find your answer ; you may actually be able to use the Henderson Hasselbach equation since the relative conjugate concentrations seem to be in mid range.
 
  • #4
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Thank you so much!
I figured that it had something to do with buffers, but I couldn't see it with the NaOH.
Thank you again for your replies.
 

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