Answer to Question #222402 in Chemistry for Kaygee

Question

1.How many millimoles of solute are contained in
a.2.95ML of 0.0789M KH2PO4?
b.0.2011L of 0.0564M HgCl2?

2.Calculate the PH of a solution that contains 2.0% (w/w)NaOH and has a density of 1.022g/ML.

Answer 1

Solution:

(1):

1 mmol = 0.001 mol

Moles = Molarity × Volume

Therefore,

a)

Moles of KH₂PO₄ = (0.0789 M) × (2.95 mL) × (1 L / 1000 mL) = 0.000233 mol

Millimoles of KH₂PO₄ = (0.000233 mol) × (1 mmol / 0.001 mol) = 0.233 mmol

Millimoles of KH₂PO₄ = 0.233 mmol

b)

Moles of HgCl₂ = (0.0564 M) × (0.2011 L) = 0.01134 mol

Millimoles of HgCl₂ = (0.01134 mol) × (1 mmol / 0.001 mol) = 11.34 mmol

Millimoles of HgCl₂ = 11.34 mmol

(2):

Determine the mass of NaOH per milliliter using the mass percentage of NaOH and the density:

(0.0200) × (1.022 g / 1 mL) = 0.02044 g NaOH mL^-1

The molar mass of NaOH is 39.997 g/mol.

Therefore,

(0.02044 g NaOH / 1 mL) × (1 mol NaOH / 39.997 g NaOH) × (1000 mL / 1L) = 0.511 mol NaOH L^-1

Molarity of NaOH = C_M(NaOH) = 0.511 M

NaOH (sodium hydroxide) is a strong base which dissociates as follows:

NaOH → Na⁺ + OH^-

[OH^-] = C_M(NaOH) = 0.511 M (according to the equation above)

We can convert between [OH^-] and pOH using the following equation:

pOH = - log[OH^-]

pOH = - log(0.511) = 0.29

pOH = 0.29

For any aqueous solution at 25^∘C:

pH + pOH = 14

Thus,

pH = 14 - pOH = 14 - 0.29 = 13.71

pH = 13.71

Answers:

(1): a) 0.233 mmol KH₂PO₄; b) 11.34 mmol HgCl₂

(2): pH = 13.71