acid problem

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whathashizzyDDS

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destroyer chem: what is pH of 1x10^-4 Ca(OH)2?


i got 10.3... just wanna make sure i did it right.

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destroyer chem: what is pH of 1x10^-4 Ca(OH)2?


i got 10.3... just wanna make sure i did it right.

Got same.

find the ph of a 0.04M NaCN solution. Given that Ka of HCN is 5.0x10^-10

Weak base problem.
NaCN + H2O -> HCN + OH-

So (OH)(HCN) / (NaCN) = Kb

They give you Ka, but CN- is the base so you need to solve for the Kb. 1.0^-14/Ka = Kb = 2x10^-5

Anyway ice table...
NaCN + H2O -> HCN + OH-
I 0.4M ............ +x .... +x
C -x................. +x .... +x
E 0.4M-x ............x ..... x

so (x)(x) / (0.4-x) = 2x10^-5
you can approximate since CN- is a weak base it won't react completely. So 0.4-x = 0.4 almost
(x^2) / (0.4) = 2x10^-5
x=0.002828

-log x = pOH = 2.54
pH = 11.45
 
i think im more confused about why they gave you the ka of the product but youre able to use it in the equation like that.
 
i think im more confused about why they gave you the ka of the product but youre able to use it in the equation like that.
Because you can figure out Kb if they give you Ka.

Ka X Kb = Kw (water dissociation constant) you NEED to memorize this number. It's 1.0 X 10^-14.

So: Ka X Kb = 1.0 X 10^-14 - rearrange to find Kb

Kb = 1.0 X 10^-14 / Ka - substitute in Ka
Kb = 1.0 X 10^-14 / 5.0 X 10^-10

You can rewrite this to be 10 X 10^-15 / 5.0 X 10^-10
Solve: 2 X 10^-5

Also there's a typo..should be 0.04M not 0.4M in the solution.

So:

(x)(x) / (0.04-x) = 2x10^-5
you can approximate since CN- is a weak base it won't react completely. So 0.04-x = 0.04 almost
(x^2) / (0.04) = 2x10^-5 - Multiply by 0.04

(2.0 x 10^-5) x (4.0 x 10-2) = 8.0 x 10^-7 = x^2 - Rewrite - 80 x 10^-8 - take sqr root
Just under 9 (9^2 = 81) so 8.9ish x 10^-4 = x

So in the equation [OH-] = x = 8.9ish x 10^-4
pOH = -log (8.9 x 10^-4)
pOH = Almost a full 1 pH less than 4...round to about 3.2
pH + pOH = 14
pH = 14 - pOH so pH = 14 - 3.2 = 11.8 ish
 
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