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SQUARE ROOT SIMPLIFY EQUATIONS CALCULATOR
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HA(aq)
H3O+(aq)
A-(aq)
Initial Conc. (M)
0.500
0
0
Change in Conc. (M)
- x
+ x
+ x
Equilibirum Conc. (M)
0.500 - x
x
x

  • Substitute into the equilibrium expression.  Assume that 0.500 - x ~ 0.500.  Simplify equation and solve for the change.

    • Example Problem

    Solution to Example Problem

  • Check answer to see if it is within limits set by your instructor. (Here we use 5%.)

    • (0015/0.500) x 100 = 0.03%

    The change is only 0.03% of the initial value and is negligible.

  • Determine the equilibrium concentrations of each species

    • [H3O+] = [A-] = x = 1.5 x 10-4 M

    [HA] = 0.500 - 1.5 x 10-4 = 0.500 M

  • Check work.

    • Work Check

    Top

    K and Q Lie on Opposite Sides of One (K>>Q or K<<Q)

    When K is much larger than Q, or K is much smaller than Q, the change in amount of each species will be very large as the system moves towards a state of equilibirum.  When this occurs, finding the change in concentration can often be facilitated by doing the following:

    • When K>>Q and K > 1, assume 100% conversion into products, followed by the back reaction to establish equilibrium.  When K<<Q and K < 1, assume 100% conversion into reactants, followed by the forward reaction to establish equilibrium.
    • Make an ICE chart to determine change and equilibrium quantities starting with those resulting from the 100% conversion.
    • Substitute quantities into the equilibrium expression.
    • Assume the change is near zero such that "[A] - x" is equal to "[A]."
    • Solve for the variable.
    • Check to see if the change is less than 5% of the maximum amount, or within the limits set by your instructor. If not, use the method of approximations, a programmable calculator, or other method to solve.
    • Solve for the equilibrium concentrations if asked to do so.
    • Check your work.
    Example: An evacuated flask is filled with sufficient H2 and I2 gases so that the concentration of each gas is 0.620 M. It is then heated to 298 K. What is the concentration of each species when equilbrium is established?

    H2(g) + I2(g) double arrows 2 HI(g) Kc = 794 @ 298 K

    • Initially the [HI] = 0, so K >>Q  and K is > 1. The change in the concentration of each species will be large so we calculate the quantity of product formed assuming 100% conversion.

      • 100% conversion will result in the formation of 1.24 M HI (1 to 1 to 2 proporation) with neither reactant remaining.

    • Make an ICE chart starting with the concentrations after the 100% conversion.
      • H2
      I2
      HI
      Initial Concentration (M)
      0
      0
      1.24
      Change in Concentration (M)
      + x
      + x
      - 2 x
      Equilibrium Concentration (M)
      0 + x
      0 + x
      1.24 - 2 x

       

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