Question

Calculate the solubility (in g / L) of a generic salt with a formula of A2B, a Ksp of 5.50 × 10−14 and a molar mass of 135 g / mol.

___ g/L

Tris [tris(hydroxymethyl)aminomethane] is a common buffer for studying biological systems.

(K_a = 5.01

×

10⁻⁹

and

pK_a = 8.3)

(a) Calculate the pH of the Tris buffer after mixing 13.0 mL of 0.10 M HCl solution with 25.0 mL of 0.10 M Tris.

(b) This buffer was used to study an enzyme-catalyzed reaction. As a result of the reaction, 0.00020 mol of H⁺ was produced. What is the pH of the buffer at the end of the reaction?

(c) What would be the final pH if no buffer were present?

Answer 1

1) You have the expression of Ksp:

Ksp = [A +] ^ 2 * [B-] = (2 * X) ^ 2 * X = 4 * X ^ 3

X is cleared = (Ksp / 4) ^ (1/3) = (5.5x10 ^ -14 / 4) ^ (1/3) = 2.4x10 ^ -5 M

It is calculated in g / L: 2.4x10 ^ -5 mol / L * (135 g / 1 mol) = 3.24x10 ^ -3 g / mol

2) a) The moles of Tris and HCl added are calculated:

n Tris = M * V = 0.1 M * 0.025 L = 0.0025 mol

n HCl = 0.1 M * 0.013 L = 0.0013 mol

The HCl reacts with Tris (decreasing it) and forms Tris + (increasing it), the pH is calculated with the equation of henderson hasselbach:

pH = pKa + log (n Tris / n Tris +) = 8.3 + log (0.0025 - 0.0013 / 0.0013) = 8.27

b) The H + reacts with the Tris (decreasing it) and forms Tris + (increasing it), the pH is calculated with the equation of henderson hasselbach:

pH = 8.3 + log (0.0012 - 0.0002 / 0.0013 + 0.0002) = 8.12

c) The concentration of H + in 38 mL of solution is calculated:

[H +] = n / V = 0.0002 / 0.038 = 0.005 M

pH = - log [H +] = - log 0.005 = 2.30

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