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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 = 0.20-x

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?

  • chem -

    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 = 0.20-x

    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).

  • chem -

    I redid it and I got 11.25 :) Thanks :)

  • biology -

    Why are antibiotics and vitamins not normally autoclaved when preparing microbiological media?

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