Question

Use the References to access importaat values if seeded An aqueous solution is prepared to be 0.290 M in sodium nitrite and 0.151 M in acetic acid (1) Is this solution a buffer solution? (2) What is the pH of this solution? pH 3) Ir0.104 moles of perchloric acid are added to one liter of this solution, what is the pH of the resulting solution? p Use the Tables link in the References for any equilibrium constants that are required

Answer 1

1) A buffer solution is made up of an acid and its conjugate base, in this case we are NOT in the presence of a buffer, since the salt is not the conjugate pair of the acid.

2) The hydrolysis reaction of the acid is:

HA + H2O = A- + H3O +

From the expression of Ka we have:

Ka = [A-] * [H3O +] / [HA]

We replace:

1.8x10 ^ -5 = X ^ 2 / 0.151 - X

We cleared:

X ^ 2 + 1.8x10 ^ -5 X - 2.72x10 ^ -6 = 0

We apply equation of the second degree and:

X = [H3O +] = 1.64x10 ^ -3 M

Since the dissociation of the nitrite ion is very low, it is disregarded and pH is calculated:

pH = - Log [H3O +] = - Log (1.64x10 ^ -3) = 2.79

3) If 0.104 moles of perchloric acid are added to one liter of solution, it is assumed that a concentration of hydronium ion of 0.104 M is added, the new concentration of hydronium and the pH are calculated:

[H3O +] = 1.64x10 ^ -3 + 0.104 = 0.106 M

pH = - Log (0.106) = 0.975

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