☰ Contents · Chemistry (General chemistry)

Normal concentration

Lessons 19 · 1 lessons · S. Masharipov, A. Mutalibov, E. Murodov, H. Islomova. General chemistry, Grade 11, 1st edition. G‘afur G‘ulom Publishing and Printing House, Tashkent, 2018
19

Normal concentration

Textbook: pp. 88–92
GoalExplain the equivalent amount and normal concentration (C_N) and solve problems with n_eq = m/E and C_N = n_eq/V.
New words
equivalent amount · ekvivalent miqdornormal concentration · normal konsentratsiyag/equiv · g/ekvneutralisation · neytrallanish
Explanation

The equivalent amount of a solute n_eq is found by dividing its mass m by its equivalent mass E: n_eq = m/E (the book uses g/equiv). The number of equivalents of solute in 1 L of solution is the normal concentration C_N of the solution: C_N = n_eq/V, with the unit N («normal»); from it n_eq = C_N·V and V = n_eq/C_N. Normal concentration depends on the equivalent mass of the substance, that is on the type of reaction: 1 M H₂SO₄ is 2 N (E = 98/2 = 49), 1 M HCl is 1 N, 1 M H₃PO₄ is 3 N for complete neutralisation. By the law of equivalents substances react in equal numbers of equivalents, so in neutralisation C_N₁·V₁ = C_N₂·V₂; for dilution C₁V₁ = C₂V₂ also holds. Note: «normality» is no longer recommended in current international practice, but it is used in school and test problems.

Worked examples
250 mL of solution contains 20 g of NaOH (E = 40). n_eq = 20/40 = 0.5; C_N = 0.5/0.25 = 2 N.
Molar concentration of 2 N H₂SO₄: z = 2, C_M = C_N/z = 2/2 = 1 M. Conversely, in 0.3 M Al₂(SO₄)₃ z = 6 (2 Al³⁺ · 3), so C_N = 0.3·6 = 1.8 N.
Class activity

Table: groups calculate the normal concentration of a 1 M solution of HCl, H₂SO₄, Ca(OH)₂ and Al₂(SO₄)₃ and enter it in a common table on the board.

Practice
1
How many equivalents are in 98 g of H₂SO₄ (E = 49)?
2
3 L of solution contains 6 equivalents of solute. What is C_N in N?
3
How many grams of KOH are in 2 L of 0.5 N KOH solution (E = 56)?
4
How many mL of 0.4 N HCl neutralise 20 mL of 0.5 N NaOH?