Question

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.

pKb for the weak base = 4.82.

Molar mass of the weak base = 82.0343g/mole.

Note: pKa = -logKa

pKb = -logKb

pH + pOH = 14

[H+ ] [OH- ] = 10^-14

Answer 1

11.39

Step-by-step explanation:

Given that:

`pK_(b)=4.82`

`K_(b)=10^(-4.82)=1.5136* 10^(-5)`

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (1.805\ g)/(82.0343\ g/mol)`

`Moles= 0.022\ moles`

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity=(0.022)/(0.055)`

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

`K_(b)=\frac {\left [ BH^(+) \right ]\left [ {OH}^- \right ]}{[B]}`

`1.5136* 10^(-5)=\frac {x^2}{0.4-x}`

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39