# Ka greater than one=strong acid?

#### thebillsfan

##### Unseasoned Veteran
10+ Year Member
5+ Year Member
I read this somewhere but now that I think about it this can't be correct, can it?

#### RogueUnicorn

##### rawr.
7+ Year Member
Ka for strong acids is irrelevant but yes that seems too low for strong acids as 1 means half-dissociating in water

#### kentavr

10+ Year Member
Ka for strong acids is irrelevant but yes that seems too low for strong acids as 1 means half-dissociating in water
Why Ka is irrelevant? I thought that the strong acid by definition is the acid that disassosiates completely. (is that correct?) Well, nothing is complete, and that means that Ka is just a very very big number, like 10^60 , it may not be measurable since [HA] concentration is very small. Ka=1 does not look like a big number to me #### RogueUnicorn

##### rawr.
7+ Year Member
Why Ka is irrelevant? I thought that the strong acid by definition is the acid that disassosiates completely. (is that correct?) Well, nothing is complete, and that means that Ka is just a very very big number, like 10^60 , it may not be measurable since [HA] concentration is very small. Ka=1 does not look like a big number to me your definition is correct.. i say it's irrelevant because it's so huge and serves no real purpose.

#### Charles_Carmichael

Moderator Emeritus
10+ Year Member
7+ Year Member
Yup, as an acid becomes stronger, it's Ka value approaches infinity (since the numerator is getting larger while the denominator is getting closer to zero). A Ka > 1 means that the acid is strong since it dissociates a significant amount compared to weak acids (10^-14 < Ka < 1). A Ka of 1 means that more than half of HA has dissociated (since x2 is in the numerator).

#### RogueUnicorn

##### rawr.
7+ Year Member
no a Ka of 1 would be exactly half - hardly acceptable under the casual "strong acid" definition

#### Charles_Carmichael

Moderator Emeritus
10+ Year Member
7+ Year Member
no a Ka of 1 would be exactly half - hardly acceptable under the casual "strong acid" definition
Hmm...isn't the equation Ka =[H+][A-]/[HA]? And at equilibrium (for example initial [HA] = 1M), [H+] = [A-] = x, while [HA] = 1 - x? So, 1 = (x^2)/(1-x). If half of it has dissociated, wouldn't the value of x be 0.5, so (0.5*0.5)/(0.5) equals 0.5, not 1?

Chemistry is not my greatest strength, but that's what I remember from my review. Am I missing something?

OP
T

#### thebillsfan

##### Unseasoned Veteran
10+ Year Member
5+ Year Member
i think he was just saying its around half, as opposed to the extreme of complete dissociation

#### RogueUnicorn

##### rawr.
7+ Year Member
i did:

2M HA --> 1M HA + 1M A- + 1M H+

/ = 1

hmm... but your math is correct as well... hmmmmmmm

#### RogueUnicorn

##### rawr.
7+ Year Member
both of us might be doing this completely wrong... i.e. changing the original concentration actually changes the ratio of [a-]:[ha] because the K value is a ratio of [a-]^2:[ha]

ah i learned somethign today... nice.

#### kentavr

10+ Year Member
What is the problem?
for HA -> H + A
x^2/(1-x) = 1 => x^2=1-x => x^2+x-1=0
x = (-1 +/- Sqrt(1+4))/2 = (-1 +/- Sqrt(5))/2
Minus doesn't works since x become <0 so
x=(-1+Sqrt(5))/2 ~ 0.6.
We have HA left 0.4
H and A produced ~ 0.6
Not exactly half but close enough.

#### RogueUnicorn

##### rawr.
7+ Year Member
given a fixed Ka, the ratio of [a] to [ha] shifts depending on originial [ha]. which makes sense. i.e. an acid that might dissociate exactly halfway at a certain concentration will not at another.l

#### kentavr

10+ Year Member
given a fixed Ka, the ratio of [a] to [ha] shifts depending on originial [ha]. which makes sense. i.e. an acid that might dissociate exactly halfway at a certain concentration will not at another.l
Agree,
Skipping the math it looks like When [HA]=2 the dissociation will be exactly halfway

7+ Year Member
word.