[seal308]

Hi, I have a question on calculating Ka or Kb.
I understand the formula and how it works.
My teacher told me a method of finding Ka or Kb.
It was to say that b/c one of the concentrations is so small. You can considered it negligible.
For instance.
Kb = x^2/0.67 - x = 2.5E-7
you can assume x in the demoninator is negligible b/c its so small.
so now have
Kb = x^2/0.67= 2.5E-7

so my question is how can you tell if the x value in the denominator is neglibile or not?
Thanks!

[sjb]

Hi, I have a question on calculating Ka or Kb.
I understand the formula and how it works.
My teacher told me a method of finding Ka or Kb.
It was to say that b/c one of the concentrations is so small. You can considered it negligible.
For instance.
Kb = x^2/0.67 - x = 2.5E-7
you can assume x in the demoninator is negligible b/c its so small.
so now have
Kb = x^2/0.67= 2.5E-7

so my question is how can you tell if the x value in the denominator is neglibile or not?
Thanks!

At a simplistic level, practice. Try calculating with and without removing the x and see what you get (though with will form a cubic, which may be harder to solve)

[Borek]

So called 5% rule - 1-0.05≈1 - so as long as x is smaller than 5% of the other value, you can ignore it.

After you finish calculations using this assumption you should check if it was valid.

[seal308]

ok so i think i get it

so i would be (x/initial conc of acid or base in the reactant) * 100
if it's less than or is 5% then x is negligible.
Is that it?

[seal308]

i'm trying this method but it doesn't seem to be working.
Here is the question i'm working on
   
Potassium nitrite is used as a food preservative. Determine the pH and concentrations of all species in an aqueous solution of KNO2 that has a concentration of 0.22 M
HNO2
5.6 × 10−4

Here is how i tried to solve it.
major species are K NO2 and water
k is spectator ion
NO2 comes from HNO2 so i know it's a base so:
NO2 + H2O  - - > HNO2 + OH^-

I do an ICE table

Initial
NO2 = 0.22M
HNO2 = 0
OH = 0

Change
NO2 = -x
HNO2 = +x
OH = +x

Equilibrium
NO2 = 0.22-x
HNO2 = +x
OH = +x

so i try this equation to find x
x^2/0.22 = 5.6 × 10−4
solve for x and i get:
x = 0.0110995495

now i check if it's within 5 %

0.0110995495/0.22 * 100 = 5.0452...
so i see that it's above 5%
so i use this equation:
x^2/0.22-x = 5.6E-4
rearrange
x^2 + 5.6E-4 - (5.6E-4*0.22) = 0
use quadratic equation get: x = 0.0108230807

apparently i'm way off b/c the answer says that OH and HNO2 are equal to 2.0E-6

Help please

[seal308]

Oh nvm i found the answer to my own problem.

HNO2
5.6 × 10−4

is the ka value of HNO2

my equation
NO2 + H2O  - - > HNO2 + OH^-

therefore i need the Kb of NO2
so i do:
Kw/Ka
1.0E-14/5.6E-4 = ans 1
square root of (ans1 * 0.22) = 1.98208...E-6

check if within 5%
and find it is well below 5% so valid.
Thanks again

[Borek]

Now try to calculate pH of 0.0001M acetic acid.

[seal308]

well i dont' know how to find it without a Ka value.
googled and found 1.8 x 10^-5
x^2/0.001 = 1.8E-5
x = 1.341640786E-4
check if 5%
1.34.../0.001 *100 = 13.41... so i think i gotta use quadratic equation

x^2/0.001-x = 1.8E-5
x^2 +1.8E-5x - (1.8E-5*0.001) = 0
do quadratic and found
x = 1.25465609E-4

however this is now within 5 % so i guess i'm doing something wrong.

I know if i want to find ph it would be -log(x) b/c x is the hydronium concentration

[Borek]

well i dont' know how to find it without a Ka value.
googled and found 1.8 x 10^-5

Good - you can't calculate pH without Ka.

x^2/0.001 = 1.8E-5
x = 1.341640786E-4
check if 5%
1.34.../0.001 *100 = 13.41... so i think i gotta use quadratic equation

OK

x^2/0.001-x = 1.8E-5
x^2 +1.8E-5x - (1.8E-5*0.001) = 0
do quadratic and found
x = 1.25465609E-4

however this is now within 5 % so i guess i'm doing something wrong.

No, it is not within 5%. Check your math.

I know if i want to find ph it would be -log(x) b/c x is the hydronium concentration

Right - and you will get correct result.

That is, I asked about pH of the 0.0001M solution and you calculated pH of the 0.001M solution, but it was about a method, and you applied it correctly 

[Sophia7X]

A good quick way to find [H+] without using an ICE table or any quadratics (only if the Ka value given is 10^-5 or smaller, as this usu. fits the 5% rule) is:
[H⁺] = sq root(Ka*[weak acid])
or [OH-] = sq root(Kb*[weak base])

[seal308]

thanks for the suggestion sophia.