Formic acid (HCO2H) has a dissociation constant of 1.8 × 10^-4 M. The acid dissociates 1:1. What is the [H+] if 0.1 mole of formic acid is dissolved in 1.0 liter of water? Use a scientific calculator.

Question

aaronq · Answer

use an ICE table and the eq. expression for the dissociation:

$K=\dfrac{[H^+][CO_2H^-]}{[HCO_2H]}$

anonymous · Answer

So the answer would equal K?

aaronq · Answer

nope the answer is equal to $[H^+]$

anonymous · Answer

Oh, so what should I do to solve the problem?

anonymous · Answer

I am not sure I understand.

aaronq · Answer

make an I.C.E. table (initial, change, equilibrium), plug your values into the equilibrium expression to find $[H^+]$.

anonymous · Answer

Do I use the equation you put up at the top?

aaronq · Answer

yes, plug your values from the ICE table into it (the values from E).

anonymous · Answer

I am sorry, but I am confused on what numbers to plug into where.

aaronq · Answer

have you made the table?

anonymous · Answer

I wrote out the formula you gave me, but I have never done this before.

aaronq · Answer

you need to make a table with the concentrations, 0.1 moles in 1 L

first the initial concentration:    $Molarity=\dfrac{0.1~mol}{1~L}=0.1~ M$

                      $HCO_2H \rightarrow H^++HCO_2^-$
Initial                0.1              0             0
Change             -x              +x          +x
Equilibrium        0.1-x          x              x


so now plug the values from the equilibrium line (last line) into the equilibrium expression.
Solve for x.

NOTE: you can ignore x from 0.1-x since the K is really small. You can double check this by verifying that x is less than 5% of the original (i.e. 0.1).

anonymous · Answer

Thank you so much for your help! My teacher just sent me an email saying for me to skip this question because we have not gone through it in class yet. I appreciate your help very very much!! :)

aaronq · Answer

haha no problem !