Question

Based on the data on the first page, how do you calculate H3O+, the concentration of molecular form of acid or base at equilibrium and the value of Ka or Kb?

Report Data Part I: Determination of Ka or Kb for an UninotiT solution Unknown number: Concentration of unknown: ON (0.100 M, unless you are told otherwise) pH: S.29 Part II: Preparation of a Buffer Solution Target pH for the buffer 568 Perform calculations related to the preparation of the buffer here: o. 3 s- log (H

Answer 1

Answer : Calculations for part (I) shown.

1) Calculation of [H₃O⁺] :

Formation of [H₃O⁺] species is as ,

H₂O + H⁺ (aq) -----> H₃O⁺ (aq)

Hence, [H₃O⁺] can be calculate using pH value of thesolution.

pH is defined as,

pH = -log[H₃O⁺]

$\Rightarrow$ [H₃O⁺] = 10^-pH.

We are given with pH = 8.29 (Basic medium) (Part I)

$\therefore$ [H₃O⁺] = 10^-8.29.

$\therefore$ [H₃O⁺] = 5.13 x 10^-9 M.

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2) Calculation of concentration of Molecular form of Acid / Base.

Any acid of the form HA ionizes by following equation,

HA (aq) + H₂O <------> H₃O⁺ (aq) + A^- (aq).

From reaction stoichiometry its clear that

[H₃O⁺] = [A^-] = 5.13 x 10^-9 M ------------ (For Part I)

And Before ionization initially [HA] = 0.100 M.

On ionization,

[HA] = 0.100 - (5.13 x 10^-9) = 0.100 M (Approximately).

3) Expression for Ka of given acid is given as,

Ka = [H₃O⁺][A^-] / [HA]

On substituting values in above equation.

Ka = [5.13 x 10^-9][5.13 x 10^-9] / 0.100

Ka = 2.63 x 10^-16 .

This is how value of Ka is calculated.

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