Calculate the percent ionization of a 0.20 M HCN solution. (Ka- 4.9x10-10) Calcu

Question

Calculate the percent ionization of a 0.20 M HCN solution. (Ka- 4.9x10-10) Calculate the percent ionization of 0.50 M triethylamine. ((CH3)3N) (Kb-6.4x10-5) Calculate the pH of a 2.00 L solution made of 0.60 M acetic acid and 0.60 M of potassium acetate. (Ka=1.8x10-5) The volume of an adult's stomach ranges from 50.0 mL when empty to 1.00 L when full. If the stomach has a volume of 400.0 mL and it's contents have a pH = 2.00, how many moles of H+ does the stomach contain? Assuming all of the H+ comes from HCl, how many grams of sodium hydrogen carbonate (NaHCO3 or baking soda) are needed to neutralize the stomach acid?

Explanation / Answer

1) Solution

The dissociation equation of hydrocyanic acid, HCN will be given as:
HCN < ---------> H+ + CN–
Let us assume ? is the degree of ionization of HCN
Initial concentration:
[HCN] = 0.2
[H+] = 0
[CN–] = 0
Equilibrium concentration:
[HCN] = 0.2 (1 – ?)
H+] = 0.2 ?
[CN–] = 0.2 ?
Ka = [H+] [CN–] / [HCN]
4.9 x 10-10 = (0.2 ?) (0.2 ?) / 0.2 (1 – ?)
As, 1 – ? ? 1
Therefore, 4.9 x 10-10 = (0.2 ?) 2
? = (4.9 x 10-10 / 0.2)1/2
= 4.95 x 10-5
Hence, percent ionization = 0.00495 %

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