GenChem Question

[swolly]

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Hey guys, I just had a general question about gen chem... I was wondering for questions where it involves calculating K(eq) or Q or K(sp) or any other questions where you write out products over reactants, when do you actually use coefficients?? Like for example if the equation was 2A + B --> 2C, sometimes it tell me to use (2C)^2/(2A)^2 x B, but in other questions it leaves out the coefficients and only uses the powers. I should probably know this, but it keeps getting me stuck. Please respond if you know what I'm talking about!! Thanks guys.

 

[Mufasha]

With the equilibrium constant expressions, you'll need to put the coefficient as an exponent on the concentrations of products and/or reactants (you won't see reactant concentrations with a Ksp, the only the products). Note that you won't ever see the coefficient as a coefficient in the actual expression (i.e., the Kc for 2A + B --> 2C will be [C]^2 / [A]^2 * ). I think you're getting mixed up with what happens in problems that require input into this expression, like a common ion problem.

If you are doing a problem that involves making an ICE chart (i.e., a common ion problem), you'll need to include coefficients as coefficients (NOT as exponents) to get each product/reactants' expression for molar solubility. For example, this is where you'll see a term like 2x if your initial concentration is 0 and the coefficient is 2 for a given product/reactant. This could then be plugged into the equilibrium constant expression to solve for molarity solubility or the Kc (depending on what else you're given and what the question is asking for). Does that make sense?

 

[Project Pat]

You also need to remember that pure solids and liquids are NEVER included in the equilibrium expression. If you see a reactant followed by (l) or (s) don't include it.

 

[swolly]

I kind of understand... so generally I should use coefficients in common ion problems and otherwise when solving for q or k without a common ion already present just use the exponents?? thanks a bunch for the help guys, just trying to sort things out.

 

[Project Pat]

You need to remember to balance the equation when dealing with these types of problems. For example: "Calculate Keq for the synthesis of ammonia at 400°C if the following concentrations are present at equilibrium: N2(g) = 1.2 mol/L, H2(g) = 0.80 mol/L, and NH3(g) = 0.28 mol/L.
N2 + 3H2 ↔ 2NH3"

1. Balance the equation: N2 + 3H2 ↔ 2NH3 - this is balanced.

2. Keq = [products]^coefficient / [reactants]^coefficient. It this case it would be: [0.28]^2 / [1.2]^1 * [0.80]^3 = 0.13

But if you are given something along the lines of: AgCl (s) = Ag+(aq) + Cl¯(aq)

AgCl(s) would NOT be included in the expression because it is a pure solid, so the final answer would be Keq = [Ag+] [Cl¯]

I hope that helped.

 

[Mufasha]

I kind of understand... so generally I should use coefficients in common ion problems and otherwise when solving for q or k without a common ion already present just use the exponents?? thanks a bunch for the help guys, just trying to sort things out.

Where are you in your studies? If you've covered Chad's already, I'll take a look at the pertinent videos again and take some notes. I also found that the Destroyer problems relating to this will help clarify the concepts.

 

[swolly]

Yea I'm good on leaving solids out of the equations. Ive watched chads and taken notes, been working through destroyer and bootcamp for orgo and GC. Ok.. so like in chads example of solving for Ksp of BiI3 given molar solubility 1.32 x 10^-5... the ksp ends up being equal to 27x^4
((x)((3x)^3)). then plug in molar solubility for x and find ksp.. but then on precipitation of Mg(oh)2 example 2 problems later, ksp= (mg2+)(OH)^2... So why wouldn't this be ksp = (mg2+)(2OH)^2 ??? I'm sorry guys! I don't know why this isn't clicking for me.

 

[elhammy]

Yea I'm good on leaving solids out of the equations. Ive watched chads and taken notes, been working through destroyer and bootcamp for orgo and GC. Ok.. so like in chads example of solving for Ksp of BiI3 given molar solubility 1.32 x 10^-5... the ksp ends up being equal to 27x^4
((x)((3x)^3)). then plug in molar solubility for x and find ksp.. but then on precipitation of Mg(oh)2 example 2 problems later, ksp= (mg2+)(OH)^2... So why wouldn't this be ksp = (mg2+)(2OH)^2 ??? I'm sorry guys! I don't know why this isn't clicking for me.

KSP of Mg(OH)2 would be equal to what you stated "(Mg+2)*(OH-)^2. However, when you are SOLVING for either KSP or molar solubility, you create an ICE table. Basically the ICE table method is Initial subtracted from the Change and then summed to find the Equilibrium concentration of each soluble ion in the reaction mixture. Because you have 2OH's, your change in the INITIAL concentration of OH- would be (+2x) because you are creating the product. Let the initial concentration equal zero, since YOU DO NOT HAVE ANY OH- in the start. Therefore, you have (2x)^2 OH ions in your equilibrium expression. You wouldn't express KSP as "ksp = (mg2+)(2OH)^2" because the coefficients of the equilibrium expression only dictate the exponential concentration of the products.

I really hope this made some sense. Basically when you solve, you set up the ICE table and THAT is where you get whatever coefficient is in the balanced net ionic equation (TIMES x) to the power of the same coefficient.

 

[swolly]

KSP of Mg(OH)2 would be equal to what you stated "(Mg+2)*(OH-)^2. However, when you are SOLVING for either KSP or molar solubility, you create an ICE table. Basically the ICE table method is Initial subtracted from the Change and then summed to find the Equilibrium concentration of each soluble ion in the reaction mixture. Because you have 2OH's, your change in the INITIAL concentration of OH- would be (+2x) because you are creating the product. Let the initial concentration equal zero, since YOU DO NOT HAVE ANY OH- in the start. Therefore, you have (2x)^2 OH ions in your equilibrium expression. You wouldn't express KSP as "ksp = (mg2+)(2OH)^2" because the coefficients of the equilibrium expression only dictate the exponential concentration of the products.

I really hope this made some sense. Basically when you solve, you set up the ICE table and THAT is where you get whatever coefficient is in the balanced net ionic equation (TIMES x) to the power of the same coefficient.

]

That helped.. appreciate it man... ok.. to be clear and make things simple... is it safe to say only when solving for molar solubility (x) given ksp or solving for ksp given molar solubility(x) that I use coefficients and exponents?? but otherwise if given concentrations of an ion I just use the exponents and plug in concentration values??

 

[elhammy]

Yes, but you have to be careful. If they tell you in a solution of MgOH2, if you have 2.0 molar OH ions.... that means you have 1.0 molar Mg ions, because you have 2 OH's for every Mg. The value for x, therefore, would be 1.0M not 2.0M.

Here is a link given an ice table problem. Don't confuse yourself over something as small as this. You're trying to find something different between 2 topics that are exactly the same.
Ignore the part about the quadratic formula because I haven't ever heard of anyone having to use it to find an equilibrium concentration on the DAT.

 

[swolly]

lol you guys are awesome!!