1 / 22

Warmup (5 min)

Solve: log(3.9 x 10 -4 ) = Name the compound HI and show how it dissociates in solution. How do acids and bases differ? Write down WHATEVER you remember. Warmup (5 min). Acids, Bases, and pH. Acid or Base?. Both Base Acid Base Acid Both Both Both Both. Aqueous solution

bryony
Download Presentation

Warmup (5 min)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solve: log(3.9 x 10-4) = • Name the compound HI and show how it dissociates in solution. • How do acids and bases differ? Write down WHATEVER you remember. Warmup (5 min)

  2. Acids, Bases, and pH

  3. Acid or Base? • Both • Base • Acid • Base • Acid • Both • Both • Both • Both • Aqueous solution • Have a high pH (above 7-14) • Have a low pH (below 0-7) • Feel slippery • Taste sour • Conduct electricity • Changes the color of pH paper • Corrosive: can damage skin • Neutralize before disposal

  4. Bases: produce OH- in solution OR accept H+ from compounds Mg(OH)2 KOH CaCO3 NaHCO3 NH3 Any anion Acids: produce or donate H+ ions in water (forming H3O+) HNO3 HCl H2SO4 H3C6H5O7 H3PO4 HClO

  5. Water molecules undergo autoionization and split into ions spontaneously: HOH + HOH  H3O+ + OH- Count each particle and identify each solution as neutral, basic, or acidic Neutral solution: mostly H2O, [H3O+] = [OH-] H2O OH- H2OH2O H3O+ H2O H2O OH- H2O H3O+ H2O H2O H3O+ H2O OH- H2O OH- H2OH2O H3O+ H2O H2O OH- H2O H3O+ H2O H2O H3O+ H2O OH- Acidic solution: mostly H2O, [H3O+] > [OH-] Basic solution: mostly H2O, [H3O+] < [OH-] H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O H3O+ H2O H2O H3O+ H2O H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O H2OH2O H2O H3O+ H2O H2O OH- H2OH2O H3O+ H2O H2O H3O+ H2O OH- H2O H2O OH- H2O H2O OH- H2OH2O H3O+ H2O H2OOH- H2O H2OH2O H2O H3O+ H2O

  6. The pH scale: based on [H+] or [H3O+] • Water (pH = 7) is neutral Strong Acids& Bases: (far away from pH 7) NaOH → Na+ + OH- HCl → H+ + Cl- - completely dissociate in water Weak Acids& Bases:(closer to pH 7) - partially dissociate in water

  7. Naming and Dissociation Practice:Fill in the blanks! Formula HI HClO2 HC2H3O2 H2Se HClO3 Name hydroiodic acid chlorous acid acetic acid hydroselenic acid chloric acid • Dissociation •  H+ + I- •  H+ + ClO2- • H+ + C2H3O2– •  H+ + HSe- •  H+ + ClO3-

  8. Neutralization If the acid and base are both strong, the net ionic equation will be the same every time. acid + base = salt + water Write the complete ion equation and the net ionic equation for the neutralization of nitric acid (strong acid) by sodium hydroxide (strong base). HNO3(aq) + NaOH(aq)  H2O(l) + NaNO3(aq) H++ NO3- + Na+ + OH- H2O(l) + Na+ + NO3- Get rid of everything that doesn’t change phase or compound from one side to another! H+ (aq) + OH-(aq)  H2O(l)

  9. Write the molecular and net ionic equation when lithium hydroxide (strong base) is mixed with carbonic acid (weak acid). LiOH(aq) + H2CO3(aq) H2O(l) + Li2CO3 (aq) Leave weak or insoluble things together. Separate strong or soluble things. Li+ + OH- + H2CO3  H2O + Li+ + CO32- OH-(aq) + H2CO3(aq)  H2O(l) + CO32-(aq) 2 2 2 2

  10. 2 Write the molecular and net ionic equations when magnesium hydroxide(weak) is mixed with chlorous acid (weak). Mg(OH)2(aq) + HClO2(aq)  H2O(l) + Mg(ClO2)2(aq) *net ionic is the same! Write the molecular and net ionic equations when aluminum hydroxide(weak) is mixed with sulfuric acid (strong). Al(OH)3(aq) + H2SO4(aq)  Al2(SO4)3(aq) + H2O(l) Al(OH)3(aq) + H+(aq) Al3+(aq) + H2O(l) 2 3 2 6 3 3

  11. HYDROCATCH The ball represents a (H+)

  12. Conjugates Write the conjugate acid of F- H+ + F- ↔ HF Conjugate acid: formed after a reaction when a base gains a H+ Write the conjugate base of H2SO4 Conjugate base: formed after a reaction when an acid loses a H+ H2SO4 ↔ HSO4-+ H+

  13. lost H+ gained H+ Let’s identify the acid and base, and their conjugates in the products HCl + H2O  H3O+ + Cl- Conjugate acid Base Acid Conjugate base

  14. lost H+ gained H+ NH3 + HOH  NH4+ + OH- Conjugate acid Base Acid Conjugate base

  15. lost H+ gained H+ H2CO3 + H2O HCO3- + H3O+ Conjugate acid Base Conjugate base Acid

  16. -log 3.000 = -0.4771 -log(3.9 x 10-4) = 3.4 10-5.6 = 2.5 x 10-6 Calculating pH*let’s practice some logs… Formulas : pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00

  17. 1. A tap water sample is contaminated with acid! If the [H3O+] in the sample is 8.90 x 10-3 M, calculate the pH of the water. Remember that [H3O+] is the same thing as the [H+] pH = -log[H3O+] pH = -log(8.90 x 10-3) pH = 2.05 (pH has no units) pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00

  18. 2. Most tap water samples are slightly basic. If the pH of a sample = 7.9, calculate the [H+] in the sample. pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00 [H+] = 10-pH [H+] = 10-7.9 [H+] = 1.3 x 10-8 M Units for concentration are in M, “molar”

  19. 3. What is the [OH-] of a solution that has a pOH of 3.00? [OH-] = 10-pOH [OH-] = 10-3.00 [OH-] = 1.00 x 10-3 M 4. What is the pH of this solution? pOH + pH = 14.00 3.00 + pH = 14.00 pH = 11.00 pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00

  20. 5. Calculate the [H3O+] AND [OH-] of human blood (pH =7.40) [H3O+] = 10-pH [H3O+] = 10-7.40 [H3O+] = 3.98 x 10-8 M To find [OH-], use pOH + pH = 14.00 pOH + 7.40 = 14.00 pOH = 6.60 then [OH-] = 10-pOH [OH-] = 10-6.60 [OH-] = 2.51 x 10-7 M pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00

  21. 6. What is the [H+] of a solution that has a [OH-] of 9.35 x 10-4 M? pOH = -log[OH-] pOH = -log(9.35 x 10-4) pOH = 3.03 pOH + pH = 14.00 pH = 10.97 [H+] = 10-pH [H+] = 10-10.97 [H+]= 1.07 x 10-11 M pH = -log[H+] pOH = -log[OH-] [H+] = 10-pH [OH-] = 10-pOH pOH + pH = 14.00

  22. Start on Problem Patent prelab. You may begin brainstorming with your group.

More Related