{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CHEM 2B Titration+Flowchart

CHEM 2B Titration+Flowchart - 3 O pOH =-log[OH pH pOH =...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Enderle – Spring 2006 Solving Titration Problems Identify the type: SASB, SBSA, WASB, or WBSA Find mmol of A & B. Which one is in excess? SA or SB ( SA or SB region ) Solve using any combo: pH = -log [H
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 O + ] pOH = -log[OH-] pH + pOH = 14.00 WA or WB ( buffer region ) Solve using H-H mmol A = mmol B (equivalence point ) SASB or SBSA: pH = 7 WASB or WBSA: Use ICE No titrant added...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online