Homework 8-solutions

# Homework 8-solutions - phan (dtp377) Homework 8 sutcliffe...

This preview shows pages 1–3. Sign up to view the full content.

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

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

Unformatted text preview: phan (dtp377) Homework 8 sutcliffe (51045) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 120 mL portion of 0 . 4 M acetic acid is be- ing titrated with 0 . 4 M NaOH solution. What is the pH of the solution after 110 mL of the NaOH solution has been added? The ioniza- tion constant of acetic acid is 1 . 8 10- 5 . 1. pH = 4 . 74 2. pH = 4 . 29 3. pH = 6 . 13 4. pH = 5 . 79 correct 5. pH = 6 . 52 Explanation: V CH 3 COOH = 120 mL [CH 3 COOH] = 0 . 4 M V NaOH = 110 mL [NaOH] = 0 . 4 M K a = 1 . 8 10- 5 For CH 3 COOH, (0 . 4 M)(0 . 12 L) = 0 . 048 mol For NaOH, (0 . 4 M)(0 . 11 L) = 0 . 044 mol CH 3 COOH + NaOH NaCH 3 COO + H 2 O . 048 mol . 044 mol 0 mol- . 044 mol- . 044 mol +0 . 044 mol . 004 mol 0 mol . 044 mol CH 3 COOH CH 3 COO- + H + . 004 mol . 044 mol . 23 L . 23 L . 0173913 M . 191304 M x Thus K a = bracketleftbig CH 3 COO- bracketrightbigbracketleftbig H + bracketrightbig [CH 3 COOH] 1 . 8 10- 5 = . 191304 x . 0173913 x = bracketleftbig H + bracketrightbig = K a [CH 3 COOH] [CH 3 COO- ] = ( 1 . 8 10- 5 ) (0 . 0173913) . 191304 = 1 . 63636 10- 6 Thus pH =- log bracketleftbig H + bracketrightbig = 5 . 78612 002 10.0 points What is the final volume when 100 mL of 0.200 M acetic acid solution is titrated to the equivalence point with 0.0400 M Ba(OH) 2 ? 1. 110 mL 2. 350 mL correct 3. 250 mL 4. 140 mL 5. 600 mL Explanation: [CH 3 COOH] = 0.2 M V CH 3 COOH = 100 mL [Ba(OH) 2 ] = 0.04 M Initially: n CH 3 COOH = (100 mL)(0 . 2 M) = 20 . 0 mmol Ba(OH) 2 +2 CH 3 COOH Ba 2+ +2 CH 3 COO- + 2 H 2 O 20 20- 20- 20 20 20 20 20 100 mL CH 3 COOH(aq) . 2 mol CH 3 COOH 1000 mL 1 mol Ba(OH) 2 2 mol CH 3 COOH 1000 mL Ba(OH) 2 . 04 mol Ba(OH) 2 = 250 mL Ba(OH) 2 (aq) Total volume = 350 mL 003 10.0 points A 100 ml sample of 0 . 200 M NH 3 solution is titrated to the equivalence point with 50 mL of 0 . 400 M HCl. What is the final [H 3 O + ]? The ionization constant of NH 3 is 1 . 8 10- 5 . 1. 8 . 61 10- 6 M correct phan (dtp377) Homework 8 sutcliffe (51045) 2 2. 1 . 00 10- 7 M 3. 1 . 05 10- 5 M 4. 3 . 70 10- 11 M 5. 1 . 55 10- 3 M 6. 6 . 09 10- 6 M Explanation: K w = 1 10- 14 K b = 1 . 8 10- 5 NH 3 + H + NH + 4 At the equivalence point of a titration, all of the NH 3 has reacted to form NH + 4 . Ini- tially, there were 0.02 moles of NH 3 , so at the equivalence point there are 0.02 moles of NH + 4 . The pH of the solution depends only on the weak acid now, so use the weak acid equation. The total new volume is 150 mL, so [NH + 4 ] = . 02 mL 150 mL 1000 mL 1 L = 0 . 133333 M You also need the K a NH + 4 of K a NH + 4 = K w K b = 1 10- 14 1 . 8 10- 5 = 5 . 55556 10- 10 H + = radicalBig K a [NH + 4 ] = radicalBig (5 . 55556 10- 10 )(0 . 133333) = 8 . 60663 10- 6 M 004 10.0 points Consider the titration curve of a weak base with a strong acid...
View Full Document

## This note was uploaded on 03/26/2012 for the course CHEM 302 taught by Professor Mccord during the Fall '10 term at University of Texas at Austin.

### Page1 / 7

Homework 8-solutions - phan (dtp377) Homework 8 sutcliffe...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online