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Chem 113 Week 10

by: Caroline Hurlbut

Chem 113 Week 10 Chem 113

Caroline Hurlbut
GPA 3.7

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About this Document

These notes cover pH calculations at equilibrium using ICE tables, rules for neglecting x in these calculations, and an intro to polyprotic acids.
General Chemistry II
Ingrid Marie Laughman
Class Notes
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This 3 page Class Notes was uploaded by Caroline Hurlbut on Friday April 1, 2016. The Class Notes belongs to Chem 113 at Colorado State University taught by Ingrid Marie Laughman in Spring 2016. Since its upload, it has received 14 views. For similar materials see General Chemistry II in Chemistry at Colorado State University.


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Date Created: 04/01/16
pH at Equilibrium Calculations • HNO2: Ka = 4.0 x 10^-4. You mix HNO2 in water to make a 1.0 M HNO2 solution. What is the pH at equilibrium? HNO2(aq) + H2O(l) I 1.0 M C -x E 1 - x 4.0 x 10^-4 = x^2 (1 - x) • rules for neglecting x —[∆reactant] x 100 < 5% [reactant] —only neglect x for reactants • can we neglect x in the example above? —neglect x: 4.0 x 10^-4 = x^2—>x = 0.02 —test: 0.02 M x 100 = 2% 1.0 M —2% < 5% ✓ test works [H3O+] = x = 0.02 M pH = -log(0.02) = 1.70 • NH3: Kb = 1.8 x 10^-5. You mix NH3 in water to make a 0.1 M NH3 solution. What is the pH at equilibrium? NH3(aq) + H2O(l) M 0 M 0 M 1 . 0 I C -x E 0.1 - x 1.8 x 10^-5 = x^2 0.1 - x Neglect x? 1.8 x 10^-5 (0.1) = x^2 x = 0.00134 0.00134 M x 100 = 1.34% < 5% ✓ 0.1 M [OH-] = 0.00134 M pOH = -log(0.001340) = 2.87 *Given a base—>pOH pH = 14 - 2.87 = 11.13 *Since the question asks for pH, one more step is needed to find pH from pOH • You have a 0.1 M solution of HClO at a pH of 4.27. What is the Ka? HClO(aq) + H2O(l) ⁶ ClO-(aq) + H3O+(aq) M 0 M 1 . M0 0 I C -x +x +x E 0.1 - x x x [H3O+] = x = 10^-4.27 = 5.37 x 10^-5 Ka = (5.37 x 10^-5)^2 (0.1 - 5.37 x 10^-5) Ka = 2.89 x 10^-8 Intro to Polyprotic Acids • polyprotic acids - acids with more than one hydrogen • ex. H3PO4(aq) + H2O→H2PO4-(aq) + H3O+(aq) —Ka1 is huge —since H2PO4- can still lose 2 more protons, there are 3 Ka values for the dissociation of H3PO4 —Ka1 > Ka2 > Ka3 —pKa1 < pKa2 < pKa3 • ex. H2SO4(aq) + H2O(l)→HSO4-(aq) + H3O+(aq) —Ka1 is huge —HSO4- can still lose a proton: HSO4-(aq) + H2O(l)⁶SO42-(aq) + H3O+(aq) —Ka2 = 0.011 —Ka1 > Ka2 —pKa1 < pKa2 • salt from strong acid and strong base results in neutral solution at 25˚C • salt from strong base and weak acid results in basic solution —ex. NaCH3COO is combination of strong base (NaOH) and weak acid (CH3COOH) —dissociation results in Na+ and CH3COO- —Na+ + H2O⁶NaOH + H+ x —CH3COO- + H2O⁶CH3COOH + OH- ✓ • salt from strong acid and weak base results in acidic solution —ex. NH4Cl is combination of strong acid HCl and weak base (NH3) —dissociation results in NH4+ and Cl- —Cl- + H2O⁶HClO + H+ x —NH4+ + H2O⁶NH3 + H3O+ ✓


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