1 / 10

Titrations Homework Solutions Page 602

Titrations Homework Solutions Page 602. Class Example:. NaOH (aq) + CH 3 COOH (aq)  NaCH 3 COO (aq) + H 2 O (l) 0.500 mol/L 15.00mL 7.5 mL C = ? n NaOH = Cv = (0.500 mol/L)(0.0075L) = 0.00375 mol n CH 3 COOH = (0.00375 mol) (1:1 ratio)

mahsa
Download Presentation

Titrations Homework Solutions Page 602

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. TitrationsHomework SolutionsPage 602

  2. Class Example: NaOH(aq) + CH3COOH(aq) NaCH3COO(aq) + H2O(l) 0.500 mol/L15.00mL 7.5 mL C = ? nNaOH = Cv = (0.500 mol/L)(0.0075L) = 0.00375 mol nCH3COOH = (0.00375 mol) (1:1 ratio) [CH3COOH] = 0.00375 mol/0.015 L = 0.25 mol/L

  3. Practice Problems, Page 602: #17) HNO3(aq) + NaOH(aq) NaNO3(aq)+ H2O(l) 17.85 mL25.00mL 0.150 mol/L nNaOH = Cv = (0.025 mol/L)(0.150 L) = 0.00375 mol nHNO3 = 0.00375 mol 1:1 ratio [HNO3] = 0.00375 mol/0.01785 L = 0.210 mol/L

  4. Practice Problems, Page 602: #18) Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq)+ 2H2O(l) 1.015 mol/L40.00mL 1.60 mol/L nHCl = Cv = (1.60 mol/L)(0.04 L) = 0.064 mol nMg(OH)2 = 0.064 mol HCl X 1 Mg(OH)2 = 0.032 mol Mg(OH)2 2 HCl [Mg(OH)2] = 1.015 mol/L VMg(OH)2 = n = 0.032 mol = 0.0315 L = 31.5 mL c 1.015 mol/L

  5. Practice Problems, Page 602: 19a) HCl(aq) + NaOH(aq) NaCl(aq)+ H2O(l) 0.150 M25.00mL 0.135 mol/L nNaOH = Cv = (0.135 mol/L)(0.025 L) = 0.003375 mol NaOH nHCl = 0.003375 mol HCl (1:1 ratio) VHCl = n = 0.003375 mol = 0.0225 L = 22.5 mL c 0.150 mol/L

  6. Practice Problems, Page 602: b) HCl(aq) + NH3(aq) NH4Cl(aq) 0.150 M20.00mL 0.185 mol/L nNH3 = Cv = (0.185 mol/L)(0.02 L) = 0.0037 mol NaOH nHCl = 0.0037 mol (1:1 ratio) VHCl = n = 0.0037 mol = 0.02467 L = 24.7 mL c 0.150 mol/L

  7. Practice Problems, Page 602: c) 2HCl(aq) + Ca(OH)2(aq) 2H2O(l) + CaCl2(aq) 0.150 M80.00mL 0.0045 mol/L nCa(OH)2 = Cv = (0.0045 mol/L)(0.08 L) = 3.6 X 10-4 mol nHCl = 3.6 X 10-4 mol Ca(OH)2 X 2 HCl= 7.2 X 10-4 mol HCl 1 Ca(OH)2 VHCl = n = 7.2 X 10-4 mol = 0.0048 L = 4.80 mL c 0.150 mol/L

  8. Practice Problems, Page 602: 20a) NaOH(aq) + HFaq) H2O(l) + NaF(aq) 37.82mL15.00mL 0.250 mol/L nHF= Cv = (0.250 mol/L)(0.015 L) = 0.00375mol nNaOH= 0.00375 mol (1:1 ratio) [NaOH] = n = 0.00375mol = 0.09915 L  0.0992 mol/L V 0.03782 L

  9. Practice Problems, Page 602: b) 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 21.56 mL20.00mL 0.145 mol/L nH2SO4 = Cv = (0.145 mol/L)(0.02 L) = 0.0029 mol nNaOH= 0.0029 mol H2SO4 X 2 NaOH= 0.0058 mol NaOH 1 H2SO4 [NaOH] = n = 0.0058mol = 0.269 mol/L c 0.02156 L

  10. Practice Problems, Page 602: c) 3NaOH(aq) + H3PO4(aq) Na3PO4(aq) + 3H2O(l) 14.27 mL25.00mL 0.105 mol/L nH3PO4 = Cv = (0.105 mol/L)(0.025 L) = 0.00262 mol nNaOH= 0.00262 mol H3PO4 X 3 NaOH= 0.007875 mol NaOH 1 H3PO4 [NaOH] = n = 0.007875mol = 0.550 mol/L c 0.01427 L

More Related