TITRATION EQUATION

ACID-BASE NEUTRALIZATION

Strong Acids

Strong Bases

Pick any strong acid and any strong base. Then,

1. Write the neutralization reaction between them including states of matter.
2. Write the net ionic equation for the reaction.

HCl

HBr

HI

H₂SO₄

HNO₃

HClO₄

LiOH

NaOH

KOH

RbOH

CsOH

Ba(OH)₂

Sr(OH)₂

ACID-BASE NEUTRALIZATION

Notice that no matter which strong acid you picked and which strong base you picked, the chemical equation will follow this format:

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Chemical Equation:

HA(aq) + BOH(aq) 🡪 BA(aq) + HOH(l)

strong acid + strong base 🡪 salt + water

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Net Ionic Equation:

H⁺(aq) + OH^-(aq) 🡪 HOH(l)

ACID-BASE NEUTRALIZATION

Neutralization Reaction:

Strong acids reacting with strong bases always produce a salt (ionic compound) and water.

QUANTITATIVE PRACTICE 1

HBr(aq) + KOH(aq) 🡪 KBr(aq) + HOH(l)

If 45.2 mL of an HBr solution is titrated with 52.3 mL of 1.32 M KOH to the equivalence point, what is [HBr]?

5

Plan:

1. Moles of titrant (base) used.
2. Moles of analyte (acid) neutralized.
3. Molarity acid = Moles acid / Liters acid

0.0523 L KOH

x

x

= 0.069 mol HBr neutralized

= 1.53 M HBr

SHORTCUT

6

At the equivalence point for a 1:1 mole ratio:

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mol_acid = mol_base

​

M_aV_a = M_bV_b

HBr(aq) + KOH(aq) 🡪 KBr(aq) + HOH(l)

REDO QUANTITATIVE PRACTICE 1 �WITH SHORTCUT

If a 45.2 mL solution of HBr is titrated with 52.3 mL of 1.32 M KOH to the equivalence point, what is [HBr]?

M_aV_a = M_bV_b

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M_a(45.2mL) = (1.32M)(52.3mL)

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M_a = 1.53 M (same as last time!)

QUANTITATIVE PRACTICE 2

H₂SO₄(aq) + 2 NaOH(aq) 🡪 Na₂SO₄(aq) + 2 HOH(l)

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If 42.6 mL of 0.43M H₂SO₄ is used to titrate 72.4 mL of NaOH to its equivalence point, what is [NaOH]?

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Need to modify shortcut equation for 1:2 mole ratio:

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n_H+M_aV_a = n_OH-M_bV_b

n_H+ = # of protons donated by acid

n_OH- = # of OH^- in base

QUANTITATIVE PRACTICE 2

H₂SO₄(aq) + 2 NaOH(aq) 🡪 Na₂SO₄(aq) + 2 HOH(l)

​

If 42.6 mL of 0.43M H₂SO₄ is used to titrate 72.4 mL of NaOH to its equivalence point, what is [NaOH]?

​

n_H+M_aV_a = n_OH-M_bV_b

(2)(0.43M)(42.6mL) = (1)(M_b)(72.4mL)

M_b = 0.51 M