Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Phosphoric acid is a triprotic acid (Ka1 = 6.9 x 10-3, Ka2 = 6.2x 10-8, and Ka3 = 4.8 x10-13). To find the pH of a buffer composed of H2PO4- (aq) and HPO42- (aq), which pKa value would you use in the

Solution: Phosphoric acid is a triprotic acid (Ka1 = 6.9 x 10-3, Ka2 = 6.2x 10-8, and Ka3 = 4.8 x10-13). To find the pH of a buffer composed of H2PO4- (aq) and HPO42- (aq), which pKa value would you use in the

Problem

Phosphoric acid is a triprotic acid (Ka1 = 6.9 x 10-3, Ka2 = 6.2x 10-8, and Ka3 = 4.8 x10-13). To find the pH of a buffer composed of H2PO4- (aq) and HPO42- (aq), which pKa value would you use in the Henderson-Hasselbalch equation?


Calculate the pH of a buffer solution obtained by dissolving 13.0 g of KH2PO4(s) and 26.0 g of Na2HPO4(s) in water and then diluting to 1.00 L.






Solution

We’re being asked to calculate the pH of buffer composed of H2PO4-(aq) and HPO42-(aq). To calculate for pH, we’re going to use the Henderson-Hasselbalch equation:

pH=pKa+logconjugate baseweak acid


We know that

pKa=-log Ka

So first have to figure out which Ka value we’re going to use for our calculation. Phosphoric acid (H3PO4) is a weak triprotic acid, meaning it can donate three protons (H+and it will have three equilibrium reactions.

Removing one H+ per step in H3PO4:

View the complete written solution...