[chris p]

Find the PH of NaNO2 if the Ka of HNO2 is .00045

I attempted the problem and obtained 12.17. Can someone please try to do the problem and let me know what they get? Thanks in advance.

[Borek]

I just realized... You can't calculate pH without known concentration of salt. The question as you posted it (here and elsewhere) has no solution. Sorry I haven't noticed it before.

[chris p]

Is there anyway to calculate the concentration (or Molarity) of the salt using the Ka?

[Donaldson Tan]

we have to make assumption regarding the concentration of the salt.

[lemonoman]

So to answer your question no.  Concentration is never related to the Ka of a substance.  The pH of a solution is, by definition, the concentration of H₃O⁺ (or,e quivalently, H⁺), which is determined by the ceoncentration of acidic/basic substance you started with.

Another example: HCl is hydrochloric acid.  At a concentration of 0 M, the pH of water is 7.  When you have water with 1 M HCl, your pH is -log(1) = 0.  Quite the difference, just because of the concentration eh?  And the Ka certainly doesn't change.  It's a constant (for a given set of conditions i.e. temperature)

Hope this helps

[chris p]

Since there is not set way to calculate the concentration and if I use .5 randomly as the concentration, will my answer of 12.17 ph be correct? Thanks.

[Borek]

Your answer is a way too high. Show your math.

[chris p]

I tried again.

x^2/.5=Kb

-log OH = pOH

14-pOH = 8.5

Is this answer better?

[Borek]

Much better

[yl88]

let [NO2-]=x,
                 NO2- + H2O => HNO2 +OH-
initial []      x                     0         0
change []    -y                    y          y
final []        x-y                  y          y
Kb=Kw/Ka=2.2E-11=(y^2)/(x-y)
y^2 + (2.2E-11)y - (2.2E-11)x = 0
y=(-2.2E-11 + sqr(2.2E-22 - 8.8E-11x))/2
x is the concentration of NO2-, which should be provided in the question.
plug in the concentration for x, solve for y, the concentration for OH-.
then, pH=14+log(y)
if you plug in 0.5 for x, pH is 8.52
if pH is 12.17, then x must be 9844927.323...

[victor]

So to answer your question no.  Concentration is never related to the Ka of a substance.  The pH of a solution is, by definition, the concentration of H₃O⁺ (or,e quivalently, H⁺), which is determined by the ceoncentration of acidic/basic substance you started with.

Another example: HCl is hydrochloric acid.  At a concentration of 0 M, the pH of water is 7.  When you have water with 1 M HCl, your pH is -log(1) = 0.  Quite the difference, just because of the concentration eh?  And the Ka certainly doesn't change.  It's a constant (for a given set of conditions i.e. temperature)

Hope this helps

For pH is 0 for HCl 1 M....I think the calculation is like this..
[HCl] = 1 M
[H+] = 1 M
pH = -log 1 e 0
     = 1 - log1 = 1 - 0 = 1
Am I right??
Regards,
Victor

[sdekivit]

no. 10^what = 1 --> a power can only be 1 when the power equals 0. Thus log 1 x 10^0 = 0 and thus -log 1 x 10^0 = 0

[Borek]

a power can only be 1 when the power equals 0

0 = 1

Interesting result

[sdekivit]

0 = 1

Interesting result

i mean x^0 is always 1 sorry for the bad english. I meant the exponent

[victor]

Um, actually in my school I get a little bit confused with this...
can pH or pOH be in minus?
if can't please take a look at this question.
count the pH of 100 ml HCl 3M!

If I cont it in my way, it would be like this.
[HCl] = 3 M
[H+] = 1 x 3 M = 3 M
pH = -log 3 = (minus mark)!!

Please *delete me*!
Regards,
Victor