Question

A solution is labeled "0.084 M HCN." What are the following in the solution? Express your answers to three significant figures even when only two are valid. Acid-base table

[HCN] = ______ M

[H₃O¹⁺] =_______ M

[CN^1-] =_______ M

pH = ________

Answer 1

Ka of the HCN 6.2 x 10^-10

write the dissociate equation

HCN + H₂O ----> H₃O⁺ CN^-

I 0.084 0 0

C -x +x +x

E 0.084-x +x +x

Ka = [H3O⁺][CN^-] / [HCN]

6.2 x 10^-10 = [x]x]/ [0.084-x]

X² + x6.2 x 10^-10 -5.208 x 10^-11

solve the quadratic equation

x = 7.2 x 10^-6 M

equilibrium concentrations of

[HCN = 0.084-x = 0.08399 M

[H3O+] = x = 7.2 x 10^-6 M

[CN-] = x = 7.2 x 10^-6 M

pH = -log[H3O+]

= -log [ 7.2 x 10^-6]

= 5.14