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Calculate the percent ionization of a 0.381 M solution of hydrocyanic acid.

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Answer:

.00402 %

Step-by-step explanation:

hydrocyanic acid (HCN) is a weak acid with a Ka of 6.17*10^-10 so you need to do a RICE table

initial concentration of HCN is .381 and initial H+ and CN- concentration is 0.

that means the equillibrium concentration of HCN will be (.381-x) and H+ and CN- will be x

put that into the formula for Ka, so it'll be Ka = x^2/(.381-x) and then use quadratic equation to solve. You'll end up getting x = 1.5 * 10^-5.

Then to find percent ionization you solve (concentration of formed ions / initial HCN)

Which will be (1.5 * 10^-5 / .381) * 100 = .00402 % making it a very weak acid <3

User Rubinsh
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