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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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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.
•  • 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.
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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.
H2(g) + I2(g) 2 HI(g) Kc = 794 @ 298 K