Question

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH=5.5?

Answer 1

The pH of the solution is given as pH = 5.5. From this information, we can calculate the [H⁺] concentration as

pH = - log₁₀[H⁺]

or, [H⁺] = 10^-pH

or, [H⁺] = 10^-5.5 (pH is given as 5.5)

or, [H⁺] = 0.3162 * 10^-5

A prior knowledge of buffer systems or consultation with any text book shall show that this hydrogen ion concentration is close to the K_a (acid dissociation constant) acetic acid. Hence, the buffer system in question is most likely the acetic acid/ acetate buffer system.

Now, we know K_a for acetic acid is 1.8 * 10^-5.

We revert to the Henderson Hasselbach equation for finding out the relative amounts of the conjugate base and the undissociated acid.

pH = pK_a + log₁₀ [A^-]/[HA] where [A^-] is the concentration of the conjugate base and [HA] is the undissociated acid.

• log₁₀[H⁺] = - log₁₀K_a + log₁₀[A^-]/[HA]

or, log₁₀[H⁺] = log₁₀K_a + log₁₀[HA]/[A^-]

or, [H⁺] = K_a * {[HA]/[A^-]}

Now, we substitute values from above.

0.3162 * 10^-5 = 1.8 * 10^-5 * {[HA]/[A^-]}

or, 0.3162/1.8 = [HA]/[A^-]

or, 0.176 = [HA]/[A^-]

So, the ratio of the concentrations of the acid to the conjugate base would be 0.176:1.0. Now, the molar concentration of the acid is given as 0.100 M.

Hence, concentration of the conjugate base is [A^-] =

(0.100 M)/ 0.176 = 0.568 M

The final volume of the solution is given as 100 mL = 0.100 L.

So, moles of the conjugate base present in the solution is

(0.100 L) * (0.568 moles/L) = 0.0568 moles.

Now, the molar mass of acetate is 59 gm/mole.

Hence, the amount of solid acetate that must be added to the solution is (0.0568 mole) * (59 gm/mole) = 3.3512 gm.

Ans: 3.3512 gm of the conjugate base must be added to obtain a solution with pH = 5.5