Question

Thiamine hydrochloride (vitamin B1 hydrochloride, HC12- H17ON4SCl2) is a weak acid with Ka 5 3.4 3 1027. Sup- pose 3.0 3 1025 g of thiamine hydrochloride is dissolved in 1.00 L of water. Calculate the pH of the resulting solution. (Hint: This is a sufficiently dilute solution that the autoion- ization of water cannot be neglected.)

Answer 1

pH = 6.76

Step-by-step explanation:

Ka = 3.4x10⁻⁷; 3.0x10⁻⁵g

The molarity of the vitamin B1 -Molar mass: 327.268g/mol- solution is:

3.0x10⁻⁵g * (1mol / 327.268g) = 9.167x10⁻⁸M

The Ka expression is:

Ka = 3.4x10⁻⁷ = [H⁺] [B1-] / [B1]

Where B1- is the conjugate base of vitamin B1,

In equilibrium, the concentrations are:

[H⁺] = X

[B1-] = X

[B1] = 9.167x10⁻⁸M - X

Where X is reaction coordinate

Replacing:

3.4x10⁻⁷ = [X] [X] / [9.167x10⁻⁸M - X]

3.117x10⁻¹⁴ -3.4x10⁻⁷X = X²

3.117x10⁻¹⁴ -3.4x10⁻⁷X - X² = 0

Solving for X:

X = -4.15x10⁻⁷M. False solution. There is no negative concentrations.

X = 7.51x10⁻⁸M

That means from the equilibrium of vitamin B1, the [H⁺] = 7.51x10⁻⁸M

From the autoinoization of water:

[H⁺] = 1.0x10⁻⁷M

[H⁺] in the solution is:

1.0x10⁻⁷M + 7.51x10⁻⁸M = 1.751x10⁻⁸M

As pH = -log[H⁺]

pH = 6.76