fak chem hw 6 - To(st22362 Homework 6 Fakhreddine(52420 This print-out should have 20 questions Multiple-choice questions may continue on the next

fak chem hw 6 - To(st22362 Homework 6 Fakhreddine(52420...

This preview shows page 1 - 3 out of 7 pages.

To (st22362) – Homework 6 – Fakhreddine – (52420)1Thisprint-outshouldhave20questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.0011.0 pointsSome solid NH4Cl is added to a water solutionof NH3. Which of the following statements iscorrect?1.pH of the solution decreases.correct2.[H3O+]ofthesolutionremainsthesame.3.pH of the solution increases.4.[OH-] of the solution increases.Explanation:0021.0 pointsSome solid KClO2is added to a water solutionof HClO2. Which of the following statementsis correct?1.pH of the solution increases.correct2.[OH-] of the solution decreases.3.pH of the solution decreases.4.[H3O+] will remain the same.Explanation:0031.0 pointsCalculate the pH of a solution containing 0.10M CH3COOH and 0.10 M NaCH3COO.1.4.74correct2.7.123.8.274.5.785.2.89Explanation:[CH3COOH] = 0.10 MKa= 1.8×10-5[CH3COO-] = 0.10 MThisisanacetatebufferinwhich[CH3COOH] = [CH3COO-], sopH = pKa=-log(1.8×10-5)= 4.744730041.0 pointsWhat is the approximate pH of a solution inwhich the concentration of benzoic acid is 1.0M and the concentration of sodium benzoateis 0.10 M?1.3.20correct2.2.643.6.644.1.655.5.20Explanation:bracketleftbigC6H5COO-bracketrightbig= 1.0 M[C6H5COOH] = 0.1 MpH = pKa+ logparenleftBiggbracketleftbigC6H5COO-bracketrightbig[C6H5COOH]parenrightBigg=-log(6.3×10-5)+ log0.101.0= 3.200660051.0 pointsWhatisthepHofasolutionwhichis0.600 M in dimethylamine ((CH3)2NH) and0.400Mindimethylaminehydrochloride((CH3)2NH+2Cl-)?Kbfor dimethylamine =7.4×10-4.1.3.312.10.78
Background image
To (st22362) – Homework 6 – Fakhreddine – (52420)23.10.874.11.215.2.956.10.697.11.05correctExplanation:Kw= 1×10-14Ka= 0.00074[(CH3)2NH] = 0.6 M[(CH3)2NH+2] = 0.4 MKa,(CH3)2NH+2=KwKb,(CH3)2NHApplying the Henderson-Hasselbalch equa-tion,pH = pKa+ logparenleftbigg[(CH3)2NH][(CH3)2NH+2]parenrightbigg=-logparenleftbiggKwKaparenrightbigg+ logparenleftbigg[(CH3)2NH][(CH3)2NH+2]parenrightbigg=-logparenleftbigg1×10-140.00074parenrightbigg+ logparenleftbigg0.60.4parenrightbigg= 11.04530061.0 pointsA 0.50 M solution of dimethylamine (a weakbase withKb= 7.4×10-4) would be resistantto large changes of pHZ1) when a strong acid is added, but not if astrong base is added.Z2) when a strong base is added, but not if astrong acid is added.Z3) when either a strong acid or a strong baseis added.1.only Z1) is true.correct2.under neither of the circumstances Z1)and Z2)3.only Z2) is true.4.only Z3) is true.Explanation:Mdimethylamine= 0.50 MKb= 7.4×10-4(CH3)2NH + H3O+(CH3)2NH+2+ H2OThe added acid reacts with the weak base toproduce its conjugate acid.The presence insolution of appreciable amounts of the conju-gate pair produces a buffer.
Background image
Image of page 3

You've reached the end of your free preview.

Want to read all 7 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes