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Use the References to access important values if needed for this question. A 20.4 mL sample of a 0.421 M aqueous hydrocyanic
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Answer #1

Solution:

Before addition of NaOH, the pH of the solution contains only HCN.

HCN is a weak acid, it is dissociated in water as:

HCN + H2O = CN- + H3O+

0.421 M --------0.0M--------0.0M (initial)

0.421 -X-----------X-------------X (at equilibrium)

Ka = X .X / 0.421 - X

HCN is a weak acid, therefore, X will be very small in comparison of 0.421 M

Hence, Ka = X . X / 0.421

Ka for HCN = 6.2 x 10^-10

6.2 x 10^-10 = X2 / 0.421M

X2 = 2.61 x 10^-10

X = √2.61 x 10^-10

X= 1.62 x 10^-5 M

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