HCOOH(aq) <-----> H^+ (aq)+ HCOO^-(aq)
The equilibrium constant (Ka) for Reaction 3 at 25 °C is 1.80 × 10−4 mol dm−3. Calculate the equilibrium concentration of hydrogen ions if the concentration of formic acid at equilibrium is 0.00500 mol dm−3
I don't know what reaction 3 is.
........HCOOH ==> H^+ + HCOO^-
I.......0.005.....0.......0
C.......-x........x.......x
E......0.005-x....x.......x
Substitute the E line into the Ka expression and solve for x = H^+.
Sorry I should have left that out, it's just the formula
HCOOH(aq) <-----> H^+ (aq)+ HCOO^-(aq)
The solution I posted stands.
OK thanks....
In the question, and in the example in the textbook there was no example of using an ICE table. Tables make assigning values so much easier
ICE tables are the baling wire of chemistry.
To calculate the equilibrium concentration of hydrogen ions (H+) in the reaction, we can use the equilibrium constant (Ka) and the initial concentration of formic acid (HCOOH(aq)).
First, let's set up an ICE table to track the changes in concentration:
Initial concentration: 0.00500 mol dm−3 0 mol dm−3 0 mol dm−3
Change: -x +x +x
Equilibrium concentration: 0.00500 - x mol dm−3 x mol dm−3 x mol dm−3
Now, we need to use the equilibrium constant (Ka) expression, which is:
Ka = [H+][HCOO-]/[HCOOH]
Given that Ka = 1.80 × 10−4 mol dm−3 and the concentration of formic acid at equilibrium ([HCOOH]) is 0.00500 mol dm−3, we can substitute these values into the equation:
1.80 × 10−4 = (x)(x)/(0.00500 - x)
Simplifying the equation:
x² = 1.80 × 10−4 * (0.00500 - x)
x² = 0.0000009 - 1.80 × 10−4x
Rearranging the equation:
x² + 1.80 × 10−4x - 0.0000009 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
Substituting the values:
a = 1, b = 1.80 × 10−4, c = -0.0000009
x = (-(1.80 × 10−4) ± √((1.80 × 10−4)² - 4(1)(-0.0000009))) / 2(1)
Simplifying and calculating:
x ≈ 0.00111 mol dm−3 (ignoring the negative solution since it is not physically meaningful)
Therefore, the equilibrium concentration of hydrogen ions (H+) is approximately 0.00111 mol dm−3.