chem
posted by Lena on .
Calculate the pH of 0.20 M NaCN solution.
NaCN > Na+ + CN
CN + H20+ > HCN+ + OH
Initial conc. of CN = 0.20 mol/L
change = x
equillibrium = 0.20x
HCN equill. = +x
OH equill. = 1x10^7+x
Ka= 6.2 x 10^10
Kb = KW/Ka = 1x10^14/6n2 x 10^10 = 1.6 x 10^5
1.6 x 10^5 = [x][1x10^7+x]/0.2
QUADRATIC FORMULA:
x= 1.8 x 10^3
pOH = log(1.8x10^3)
= 2.73
Ph = 14  POH
=11.26
DID I DO THIS CORRECTLY?
How would I determine whether or not the CN was an acid or a base. I was thinking this. The HCN is an acid so when you take the H off of it it becomes the conjugate base. Is this also correct?

Calculate the pH of 0.20 M NaCN solution.
NaCN > Na+ + CN
CN + H20+ > HCN+ + OH
Initial conc. of CN = 0.20 mol/L
change = x
equillibrium = 0.20x
HCN equill. = +x OK to here.
OH equill. = 1x10^7+x This is also +x.
I will let you redo the math to get pH but I solve for pH and obtained 11.25.
Ka= 6.2 x 10^10
Kb = KW/Ka = 1x10^14/6n2 x 10^10 = 1.6 x 10^5
1.6 x 10^5 = [x][1x10^7+x]/0.2
QUADRATIC FORMULA:
x= 1.8 x 10^3
pOH = log(1.8x10^3)
= 2.73
Ph = 14  POH
=11.26
DID I DO THIS CORRECTLY?
How would I determine whether or not the CN was an acid or a base. I was thinking this. The HCN is an acid so when you take the H off of it it becomes the conjugate base. Is this also correct? Yes, HCN is an acid, you remove the H and the CN^ becomes the conjugate base (because it will accept a proton). 
I redid it and I got 11.25 :) Thanks :)

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