Analytical and Equilibrium Molarity

Analytical and Equilibrium Molrity
Analytical Molarity is the total number of a solute, regardless of its chemical state, in one liter of solution
• describes how a solution is prepared(recipe) 98.0 g H2SO4 dissolved in water diluted to 1.0 L.

1.0 M H2SO4

Equilibrium Molarity or species molarity is the molar concentration of a particular specie in the solution.
• requires a careful analysis on how solutes behave when it is dissolved in solvents
1.0 M H2SO4 (AM)
0.0 M H2SO4 (EM)
Species molarity due to dissociation:
[H2SO4] = 0.00 M
[HSO4] = 0.99 M
[H3O+] = 1.01 M
[SO4] = 0.01 M

Example: Analytical and Equilibrium
Solution:
Molarity
Ex. 3-4; p. 84/Skoog
Given:
m Cl3C2O2H = 285 mg.
MM Cl3C2O2H = 163.4 g/mol
V soln = 10 mL.
% ionization = 73%
Req’d: C and [C]

1.

Compute for the molar analytical concentration
CCl3C2O2H.
2. Write dissociation Equation as:
HA = H+ + ADetermine the [HA] at
73% ionization = 23% *(CCl3C2O2H.)
[A-] = 0.73 * [HA]
[H+] = [A-] based on the equation

Example: Analytical and Equilibrium
Concentration
3-19: p. 119/Skoog
Average human blood serum contains 18 mg. K+ and 365 mg Cl-. Calculate:
a) molar conc. of each species (dserum=1 g/mL)
b) pK and pCl

p-functions p-function or p- value is another methos of expressing specie concentration. pX = - log [X]

Prominent p-functions used in
Chemistry:
• pH
• pOH

Example 3-8; p. 88/Skoog
Calculate the p-value of each ion that is 2.00 x 10^-3 M in NaCl and 5.4 x 10 ^-4 in HCl.
Solution:
pH = -log [H3o+] pH= -log[5.4 x 10 ^-4 ] = 3.27

pNa = -log [2.00 x 10^-3 ] pNa = 2.669
For pCl = -log[Cl NaCl] + -log[ClHCl] pCl = 2.595

Concentration of Solutions
2. Normality (N) is the ratio between the no. of moles of solute in one equivalent per liter of solution.
• An equivalent is the ratio between Reacting units can be a: the molar mass of the substance
•Base, Acid or salt and the Avogadro’s number of the reacting units.
• The equivalent weight is the ratio
The equivalent weight is the between the mass of the given weight of the substance in grams that will furnish 1 substance and the no. of mole of the reacting unit. equivalents. Concentration of Solutions
MM subs
# eqs = -----------------# reacting units

Step 1

Step2

Base:
MM subs
# eqB = --------------------#replaceable OH-

Acid:
MM subs
# eqA = --------------------#replaceable H+
Salts:

MM subs
# eqS = --------------------Total + charge of cation Step3

Examples: (Normality)
Chem 1 Lec guide; p. 126
Mendoza

Solution:
• eqA=

Calculate the normality of a solution containing 45g. of
H2SO4 in 500 mL solution?

• eq-wt =

Given: m H2SO4 = 45 g.
V soln = 500 mL.

• N=

Req’d: N soln

Concentration of Solutions
3. Molality (m) is the ratio between the no. of moles of solute in 1 kg. of solvent.
• Used to prepare solid analytes using solid media.
• Formula:

Example: (Molality)
Ex. #2: Chem 1 lec guide:p. 124/
Mendoza

Solution:

What is the molality (m) of a solution prepared by dissolving 2.70 g CH3OH in
25.0 g H2O?
Given:
mCH3OH = 2.70 g.; MM = 32 /gmol mH2O = 25.0 g.; MM = 18 g/mol

Req’d: m m = 3.37 mols/kg.

Assignment #1:
Due Date: next meeting
Skoog:
pp. 118-119
3-13, 3-16, 3-20, 3-23

Where to place:
Short bond paper with boarders.