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Chapter 10 odd problems terminations CHEM1405.P02 Dr. Stankus Dilution of Solutions 1. What pot of (a) 12.0 M HCl is need to f are to 2.00 L of 1.00 M HCl? Mconc x Vconc = Mdil x Vdil (12.0 M) x (Vconc) = (1.00 M) x (2.00 L) Vconc=0.167 L (b) 1.04 M Na2CO3 is need to t exclusivelyy 0.500 L of 1.00 M Na2CO3? Mconc x Vconc = Mdil x Vdil (1.04 M) x (Vconc) = (1.00 M) x (0.500 L) Vconc=0.481 L 3. What volume of concentrated (18.0 M) sulfuric acid would be required to make for each one of the train? a. 1.25 L of 6.00 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (6.00 M) x (1.25 L) Vconc=0.417 L b. 575 mL of 0.100 M solution Mconc x Vconc = Mdil x Vdil (18.0 M) x (Vconc) = (0.100 M) x (0.575 L) Vconc=0.00319 L Acid-Base Titrations 5. count on the mebibyte of an HCl solution if 20.0 mL of it requires 33.2 mL of 0.150 M NaOH for neutralization. 7. Calculate the molarity of a Ca(OH)2 solution if 18.5 mL of it requires 28.2 mL of 0.0302 M HCl for neutralization. The products are CaCl2(aq) and piddle. 9. A 20-mL specimen of gastric melted is change by 25 mL of 0.10 M NaOH. What is the molarity of HCl in the fluid? Assume that all the acidity of the gastric fluid is due to HCl. 11. How many milliliters of 0.100 M H2SO4 are required to play off with 10.3 mL of 0.404 M NaHCO3?

H2SO4 (aq) + 2 NaHCO3 (aq) ? Na2SO4 (aq) + 2 H2O + 2 CO2(g) 13. How many milliliters of 0.0195 M HCl are required to titrate a. 25.00 mL of 0.0365 M KOH(aq) b. 10.00 mL of 0.0116 M Ca(OH)2(aq), c. 20.00 mL of 0.0225 M NH3(aq) pH and pOH 15. What is the pH of each of the following solutions? a. 1.0 x 10-2 M HCl pH = -log[H3O+]= -log(1.0x10-2) = 2.00 b. 1.0 x 10-4 M HNO3 pH = -log[H3O+]= -log(1.0x10-4) = 4.00 17. What is the pOH of each of the following solutions? a. 1.0 x 10-2 M NaOH pOH = -log[OH-]= -log(1.0x10-2) = 2.00 b. 1.0 x 10-3 M KOH pOH = -log[OH-]=...If you involve to get a skillful essay, order it on our website: Ordercustompaper.com

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