lecture notes2

4 changed the buffer ph by only 034 unit chem 102 d

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

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

Unformatted text preview: hem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 7. If 0.050 mL of 6.0 M NaOH is added to 400 mL of 10-5 M HCl, what is the resulting pH? The added NaOH reacts with the HCl. mol NaOH added = (6.0 mol/L)(0.000050 L) = 0.00030 mol init mol H3O+ = (10-5 mol/L)(0.400 L) = 0.000004 mol init mol mol after rxn OH- + H3O+ 6 2H2O 0.00030 0.000004 0.000296 0 The HCl is completely neutralized and you are left with a solution of a strong base. [OH-] = 0.000296 mol/0.40005 L = 7.4 x 10-4 M pOH = 3.13 Y pH = 10.87 (Notice that the pH of this unbuffered solution increased by almost 6 pH units with this small addition of base . Adding 100 times this volume of base to a buffer (Prob. 4) changed the buffer pH by only 0.35 unit.) Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 8. Characterize the solution formed (strong acid, strong, base, weak acid, weak base, buffer or neutral) when equal volumes of the following are mixed. Explain. 0.5 M NaOH + 0.5 M HC2H3O2 The following reaction...
View Full Document

This document was uploaded on 03/06/2014 for the course CHEM 102 at CSU Northridge.

Ask a homework question - tutors are online