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Understanding Acids and Bases: Key Properties and Reactions
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Explore the properties of acids (sour taste, red litmus paper, conductivity) and bases (alkalinity, salt-water production). Learn how they react with metals and each other. Dive into pH scale details.
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Understanding Acids and Bases: Key Properties and Reactions
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Presentation Transcript
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Acids and Bases
Roselyn Dooley, Tyler Schmidt, Kyle Doubleday and Deondré Robinson - Properties of Acids -Sour taste -React with active metals -Turns litmus paper Red -React with bases to produce salt and water -Conduct electricity -1- 6.9 on pH scale
- Properties of Bases -Bitter taste -Slippery feel -Turns litmus paper blue -Reacts with acids to produce salt and water -Conduct electricity -7.1 - 14 on pH scale
- Binary and Tertiary acids Binary acid- an acid that contains only two different elements: hydrogen and one of the more electronegative elements. Tertiary acid- an acid that contains hydrogen oxygen and one more element.
- Theories of Acids and Bases -Arrhenius Acid- A substance that dissociates to produce hydrogen ions in water Base- A substance that dissociates to produce hydroxide ions in water -Bronsted-Lowry Acid- Any substance that can donate H+ ions. (A proton donor) Base- Any substance that can accept H+ ions. (A proton acceptor)
- Theories of Acids and Bases Cont. -Lewis Acid- Any substance that can accept a pair of nonbonding electrons. (electron pair acceptor) Base- Any substance that can donate a pair of nonbonding electrons. (electron pair donor)
- Naming Acids Rule #1 If the negative ion in the acid ends in "ide" you name the acid "Hyrdo (stem) ic acid Ex: HCl (Chloride) would yield Hydrochloric acid Rule # 2 If the negative ion in the acid ends in "ite", you name the acid "(stem)ous acid" Ex: HNO2 (Nitrite) would yield Nitrous Acid *Use only if there is an oxygen in the chemical formula
- Naming Acids Cont. Rule # 3 If the negative ion ends in "ate", you name the acid "(stem)ic" acid. Ex: HIO4 (Periodate) would yield Periodic acid Note: The stem of Sulfur is Sulfur Also the Stem of Phosphor is phosphor
- Name that Acid! HSCN HClO2 HClO3 HBr H2SO3 H3P
- Answers Thiocyanic acid Chlorous acid Chloric acid Hydrobromic acid Sulfurous acid Hydrophosphoric acis
- pH Scale
- Six Strong Acids HCl HBr HI HNO3 H2SO4 HClO4 Everything else is considered a weak acid
- Writing Acid-base reactions in aqueous solutions General Formulas
- Now you try
- Answers HBr H+1 + Br-1 H2SO4 H+1 + SO4-2 HCN + H2O H3O+1 + CN-1 HC2H3O2 + H2O H3O+1 + C2H3O2-1
- Neutralization Reactions between Acids and Bases Neutralization- The reaction of hydronium ions and hydroxide ions to form water molecules Example equation- HCl(aq) + NaOH(aq) NaCl(aq)+ H2O(l) Note: This is basically just a double displacement reaction
- Now you try HClO4(aq)+ NaOH(aq) HBr(aq) + Ba(OH)2 (aq) ***You might need your pink sheet
- Answers HClO4(aq)+ NaOH(aq) NaClO4(aq)+ H2O(l) 2HBr (aq)+ Ba(OH)2 (aq) BaBr2 (aq) + 2H2O(l)
- Calculate Hydronium and Hydroxide
- Hydronium and Hydroxide Answers
- Calculate pH and pOH
- pH and pOH Answers
- Titration The controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration. Once the two solutions are chemically equivalent, the solution has reached the equivalence point.
- Titration Essentially, you add an acid of known molarity (concentration) to a base of unknown molarity in measured amounts to find the unknown, or vice versa. Once at the equivalence point, the unknown concentration can be calculated using known concentration and volumes.
- Walkthrough Problem 500 mL of .5 M HF titrates with 635 mL NaOH. 1) Balance the equation: HF + NaOH NaF + H2O 2) Choose method: M1V1 = M2V2 or conversions 3) (.5M) (.500 L) = (x M) (.635 L) 4) .5 M (.500 L) .635 L 5) .4 M NaOH
- Titration Calculations Ex 1) 25 mL of .3M HCl reaches an equivalence point with 75 mL of NaOH. What is the molarity of the NaOH?
- Titration Calculations 1) Balance the equation: HCl + NaOH NaCl + H2O 2) Because the mole ratios are equal (1:1), we can use the formula M1V1 = M2V2 3) (.3M) (.025 L) = (x M) (.075 L) 4) .3 M (.025 L) .075 L 5) .1 M NaOH = x M
- Titration Calculations Ex 2) 550 mL of H2SO4 of unknown concentration reaches an equivalence point with 775 mL of 2.0 M NaOH. What is the concentration of the H2SO4.
- Titration Calculations 1) Balance the equation: H2SO4 + 2NaOH Na2SO4 + 2H2O 2) Because the mole ratios are not equal (1:2), we must use conversion factors. 3) 2.0 M NaOH = (x) mols NaOH .775 L NaOH 4) 1.55 mols NaOH x 1 mol H2SO4 = .755 mols H2SO4 2 mols NaOH 5) .755 mols H2SO4 = 1.4 M H2SO4 .550 L H2SO4
- Questions for us
- Sources http://www.google.com/imgres?imgurl=http://www.dartmouth.edu/~chemlab/techniques/graphics/titration/titration10.gif&imgrefurl=http://www.dartmouth.edu/~chemlab/techniques/titration.html&h=263&w=190&sz=32&tbnid=ZR_ZcE0XRTe1TM:&tbnh=90&tbnw=65&zoom=1&usg=__Lx_UdDHNNq5SJQMUUATJ_Hg7nCc=&docid=8iiaXS9UV6QRhM&sa=X&ei=gF6tUfDeDtCJ0QHFg4DgCA&sqi=2&ved=0CF4Q9QEwBg&dur=155 http://www.elmhurst.edu/~chm/vchembook/184ph.html http://copernicusconsulting.net/designers-are-not-researchers-the-difference-between-design-and-social-research/ http://www.thechemicalblog.co.uk/what-is-a-titration/ http://dl.clackamas.edu/ch104-04/double.htm http://www.pc.maricopa.edu/data/GlobalFiles/file/chemistry/lee/Double%20displacement%20reaction.pdf https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=bZiz86EUg2v9FM&tbnid=h7bsJowJO9PdSM:&ved=0CAUQjRw&url=http%3A%2F%2Fthepickledhedgehog.com%2F2012%2F02%2F21%2Fscience-life%2F&ei=Rl-tUYjpM8aw0AGWpoCQAw&bvm=bv.47244034,d.dmQ&psig=AFQjCNH5PX-KRxbEbTHb1ihQtytljDNj1Q&ust=1370403008833830
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