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Chemical Equilibrium: Strong Acids and Bases, Study notes of Chemistry

The concept of chemical equilibrium, focusing on strong acids and bases. It explains how these substances behave in aqueous solutions and how their concentrations are related to the equilibrium constant. The document also covers the auto-ionization of water and the concept of pH.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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Download Chemical Equilibrium: Strong Acids and Bases and more Study notes Chemistry in PDF only on Docsity! -61- STRONG ACIDS AND BASES • Before we look at the behaviour of strong acids and bases we must briefly discuss THE CONCEPT OF CHEMICAL EQUILIBRIUM.... • A chemical reaction is at equilibrium when it is proceeding forward and backward at the same rate. • The concentrations of all the reactants remain constant. For the reaction: a A + b B º c C + d D • At equilibrium: Keq = • In this expression, [A] represents the concentration (usually a Molarity, or mol L—1 ) of reactant A and Keq is the Equilibrium Constant for the reaction at that temperature. • The magnitude (how large is the numerical value?) of Keq indicates the “position” of the equilibrium; if Keq is large, the equilibrium lies far to the right; there are more products than reactants. • If Keq is small, the equilibrium lies to the left; there are more reactants than products. • Solids do not appear in Keq expressions; their concentrations (/densities) are constant. -62- WATER AS A SOLVENT • As most of the solution chemistry we study is carried out in water, or in aqueous solution, we need to consider the behaviour of water............. • Water undergoes auto-ionization to a very small extent, written as the equilibrium: 2 H2O º H3O + + OH — Keq = • What is the position of this equilibrium ? • In aqueous solutions, the concentration of water, [H2O], is roughly constant ; • So we do not include the [H2O]2 term in the Keq expression giving: • This is called the Ion Product of Water. • Its numerical value at 25EC is 1.00 x 10 —14 ........... so; • As [H3O+] = [OH —] in pure water; -65- Memorize these six strong bases !! LiOH, NaOH, KOH are all monobasic. Ca(OH)2 Sr(OH)2 and Ba(OH)2 are all dibasic. • Recall that bases are substances which either: i) increase OH — ion concentration in water ii) decrease H3O+ ion concentration in water EXAMPLES: i) Soluble hydroxides of the Groups I and II metals: Sodium hydroxide, NaOH: Barium hydroxide, Ba(OH)2: • All six are considered strong bases, 100% ionized in solution. For a summary of the common strong acids and bases, see Table 4.1 in MH5, or the Tutorial Problems Manual. • Another contrast, ammonia, NH3 is a weak base that increases OH — concentration in water: NH3 + H2O º NH4 +(aq) + OH— (aq) ammonia ammonium ion -66- ii) Many metal oxides and hydroxides are insoluble in water, but react with acids: • By decreasing [H+], Mg(OH)2 is acting as a base. • Zinc oxide, ZnO will react similarly: • Barium oxide, BaO and cadmium oxide, CdO are two more metal oxides that will act as strong bases. • So, what happens if we mix an acid with a base? • IMPORTANT: If you mix any acid with any base, the result will be a neutralization reaction. • The stoichiometry is: • The reaction will proceed until all of the Limiting Reagent (either the H+ from the acid or the OH — from the base) has been used up. • If moles H+ = moles OH —............ -67- ACIDITY: The pH Scale • The quantity pH (power of hydrogen) is defined as: pH = ! log10 [H +] • It is a convenient scale for expressing the concentration of H+ (or hydronium, H3O+) ions...... • In pure water; [H+] = 1.0 x 10!7 M pH = ! log (1.0 x 10!7) = 7.00 • So this is a neutral solution with pH = 7.00 • Because of the minus sign, pH decreases as [H+] in solution increases............... pH < 7.00 [H+] > 1.0 x 10!7 M; solution is acidic pH > 7.00 [H+] < 1.0 x 10!7 M; solution is basic • We can use pOH to express [OH —] concentration: pOH = ! log10 [OH —] • And, since [H+] [OH —] = Kw , taking logs we get: log [H+] + log[OH—] = logKw or pH + pOH = pKw = 14.00 • pH ranges from 0 (acidic) to 14 (basic), pOH ranges from 14 to 0 -70- 2) A 2.50 g sample of Ba(OH)2 is dissolved in water. This solution requires 41.50 mL of HNO3 for neutralization. What is the molarity of the HNO3 ? -71- 3) An organic acid is known to be diprotic. When 0.125 g is dissolved in water, it requires 17.5 mL of 0.0893 M KOH for neutralization. What is the molar mass of the acid ? -72- DILUTION OF SOLUTIONS • When a solution is diluted (either by adding water or another solution), the concentration of that solution is decreased. • The number of moles of the solute do not change, but the volume does.... • If two solutions are mixed together, the volume of the resultant solution is always the total volume of the two original solutions. EXAMPLE: A 50.0 mL sample of a 0.150 M solution of HNO3 is diluted by adding 75.0 mL of water. Calculate the pH of the resultant solution.
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