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AlekseyPX
3 years ago
11

Determine the percent ionization of a 0.145 m hcn solution.

Chemistry
1 answer:
iragen [17]3 years ago
5 0
When Ka of HCN  = 6.2x10^-10 

              HCN       H3O+      CN-
initial    0.145          0            0
change -X              +X          +X
final     (0.145-X)     +X          +X

Ka= [H3O+][CN]- / [HCN]
by substitution:
we can assume [HCN] = 0.145
6.2x10^-10 = X*X / (0.145) by solving this equation,
X = 9.5 x 10^-6 
∴ [ H3O+] = 9.5 X 10^-6 
∴percent ionization = [H3O+]/[HCN] *100
                               = 9.5X10^-6 / 0.145 *100
                               = 0.0066 %
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Which of these, if dissolved in 1.0 l of pure water, would produce a buffer solution? which of these, if dissolved in 1.0 l of p
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Answer:

  1. (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)

Explanation:

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.

Also, a buffer solution is a solution which resists changes in pH when acid or alkali is added to it.

  1. (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄) when dissolved in 1 L H₂O will produce a buffer because NaH₂PO₄ is considered as weak acid while Na₂HPO₄ is its conjugate base. 2.
  2. (0.1 mol H₃O⁺ + 0.1 mol Cl⁻) is not a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.
  3. (0.1 mol HCl + 0.1 mol NaoH) HCL is a strong acid and NaOH is a strong base so it will not form a buffer when dissolved in water.
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So the right choice is

  1. (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)

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