The Ka of a monoprotic weak acid is 0.00245. What is the percent ionization of a 0.142 M solution of this acid?
How do you make the ice box?
Achieve Week 2 #2
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Re: Achieve Week 2 #2
You can think of the general equation for acids: HA(aq) ↽−−⇀ H+(aq)+A−(aq). The initial concentration of HA is given as 0.142 M, and the initial concentrations of H+ and A- are both 0. The change in equilibrium (from left to right) of the ICE table is going to be -x, +x, +x since you're forming product. Thus, the equilibrium concentrations are going to be 0.142 - x for [HA], and then x for [H+] and [A-]. The equation is x^2/0.0142 - x = 0.00245 (the Ka value). From then on, it's just the quadratic formula.
Hope this helps!
Hope this helps!
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Re: Achieve Week 2 #2
The reaction will be represented as HA(aq) <--> H+(aq) + A-(aq). The initial concentration of HA is the given concentration of the acid. The initial concentration of H+ and A- are 0. The change is represented as -x for HA and +x for H+ and A-. This results in the equilbirum looking like (given concentration) - x, x, x (when you go from left to right of the chart).
Re: Achieve Week 2 #2
katrinahuwang_1L wrote:You can think of the general equation for acids: HA(aq) ↽−−⇀ H+(aq)+A−(aq). The initial concentration of HA is given as 0.142 M, and the initial concentrations of H+ and A- are both 0. The change in equilibrium (from left to right) of the ICE table is going to be -x, +x, +x since you're forming product. Thus, the equilibrium concentrations are going to be 0.142 - x for [HA], and then x for [H+] and [A-]. The equation is x^2/0.0142 - x = 0.00245 (the Ka value). From then on, it's just the quadratic formula.
Hope this helps!
This was super helpful... thank you! One question: how can we be sure that there are no stoichiometric coefficients?
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Re: Achieve Week 2 #2
I think unless stated in the problem, we can assume that there are no stoichiometric coefficients. Therefore, we would proceed and write our equilibrium expression assuming stoichiometric coefficients are 1.
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