solution for hw 3 - Chem 211 Solutions to WebCT Homework...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem 211 Solutions to WebCT Homework 1.Weak Base Titration A 10.05 mL solution of 0.0550 M NaCN is diluted with 50.00 mL of water, then titrated with 0.0633 MHClO4. What is the pH, to two decimal places, at the equivalence point? Use the textbook appendix for Ka or Kb values. Titration reaction CN- + H+ HCN At the equivalence point, all we have in solution is HCN, i.e, a weak acid. So, we would use: [CN ][H ] x x Ka = = = [HCN] C HCN - x C HCN Ka for HCN from Appendix G = 6.2 10-10 What is CHCN? - + 2 2 Chem 211 Solutions to WebCT Homework Moles CN- initially = moles HCN at equivalence point = (10.05 mL)(0.0550 M) = 0.55275 mmol HCN What is the total volume at the equivalence point? VT = 10.05 mL + 50.00 mL+ Ve Equivalence point volume, Ve is: Ve = (10.05 mL)(0.0550 M ) = 8.732 mL 0.0633 M 0.55275 mmol = = 8.036 10 -3 M 10.05 + 50.00 + 8.732 mL Then , C HCN is : C HCN Then , x2 6.2 10 -10 = 8.036 10 -3 x = 2.232 10 -6 M = [H + ] pH = 5.65 ...
View Full Document

This note was uploaded on 10/23/2008 for the course CHEM 211 taught by Professor Russianteacher during the Winter '07 term at University of Alberta.

Page1 / 2

solution for hw 3 - Chem 211 Solutions to WebCT Homework...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online