This problem is in Schaum\'s Outline of College Chemistry, so I already have the

Question

This problem is in Schaum's Outline of College Chemistry, so I already have the answer. The problem though is that Schaum's doesn't provide an explanation for why they got that answer without simply stating to use the quadratic formula to solve it. Well, I certainly know how to use a quadratic formula, but how can I apply it to this equation?

(1.4x10^-3)(.0100-x)/(.0020+x), x =?

Additional Details

Yeah, KSUBa = what is stated above. Sorry about that. Really need help with solving this. This problem is in Schaum's Outline of College Chemistry, so I already have the answer. The problem though is that Schaum's doesn't provide an explanation for why they got that answer without simply stating to use the quadratic formula to solve it. Well, I certainly know how to use a quadratic formula, but how can I apply it to this equation?

(1.4x10^-3)(.0100-x)/(.0020+x), x =? Yeah, KSUBa = what is stated above. Sorry about that. Really need help with solving this.

Explanation / Answer

weak acids/bases bleh multiply both sides by (.0020+x) to get rid of the denominator... however there should be a Ka or Kb on the other side of the equal sign, no? It seems there should be more to the given because it is meaningless as shown. acid base equilibia are usually in the format Ka = [H+] [A-] / [HA] in your case you end up with a quadratic equation that you have to solve with the QF

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