0.426 g of KHP contains how many moles of KHP? If you need 0.00253 mol of NaOH to reach the equivalence point in the titration of an HCl sample, how many mol of HCl were present in the sample?

Part 3 Prepare a data table for the titration of NaOH with diluted vinegar. Include enough data columns for at least 3 titrations. PRELAB QUESTIONS 1. Calculate the molar mass of Potassium acid phthalate, KHC8H404 (KHP). You will use this number to do the calculations for part 1 once you do the titration. 2. Calculate the molar mass of acetic acid, HC₂H3O2. You will use this number to do the calculations for part 3 once you do the titration. 3. Solve the following problem: If it takes 35.04 mL of a 0.2455 M NaOH solution to neutralize 25.00 mL of an unknown HCl solution, what is the molarity of the HCl solution? Use the roadmap given under part 2 to calculate this problem. MA Chemistry 60 lab Page 110 3

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3. A 0.270 M solution of sodium formate, used in several fabric dyeing and printing processes, is prepared by dissolving it in water. A 75.0-mL sample of this sodium solution is titrated with 0.270 M hydroboric acid (HBr, strong acid) at 25°C. What is the pH of the solution after 15 mL of HBr has been added beyond the equivalence point of the titration. Sodium formate is a basic salt, with the formula HCOONA, and the acid- ionization constant of its conjugate, formic acid (HCOOH), is 1.7 x10-4 at 25°C.

Concentration of HCl: 0.05 M Volume of HCl: 4.4 ml Concentration of NaOH: 0.0088M Volume of NaOH: 25 ml

When 30 .00 mL of 0.1011 M HCl in 50 mL of deionized water is titrated against 0.09889 M NaOH, the pH increases. What is the volume (in mL) of NaOH required to reach the equivalence point and a pH of 7.00? Find the pH when the volume of NaOH added is 0.02 mL less than the volume required to reach the equivalence point. Find the pH when the volume of NaOH added is 0.01 mL less than the volume required to reach the equivalence point. Find the pH when the volume of NaOH added is 0.02 mL more than the volume required to reach the equivalence point. Comment on the significance of the changes in pH values in relation to the increments of sodium hydroxide added when going “through” the end point. Define equivalence point. For a weak base/strong acid titration, is the pH at the equivalence point <, >, or equal to 7?

Which of the following mixtures will be buffered when dissolved in a liter of water? 1 mol Calcium hydroxide and 3 mols hydroiodic acid 1 mol of sodium chloride and 2 mols of hydrochloric acid 4 mols of ammonia and 4 mols of hydrochloric acid 2 mols of hydrobromic acid and 1 mol of sodium hydroxide 2 mols of phosphoric acid and 2 mols of sodium hydroxide

Question 6 Type numbers in the boxes. 45.0 mL of o.650 M HNO, is titrated by 0.650 M KOH. aby Part 1: 10 points Calculate the pH of the acid solution before any titrant is aby Part 2: 10 points added. aby Part 3: 1 points pH = aby Part 4: 10 points Calculate the pH after 1.47 mL of o.650 M KOH is added 31 points to 45.0 mL of 0.650 M HNO,. pH = Calculate the pH after 45.00 mL of o.650 M KOH is added to 45.0 mL of o.650 ΜΗΝΟ. pH = Calculate the pH after 89.5 mL of o.650 M KOH is added to 45.0 mL of 0.650 M HNO,- pH =

A 15.5 mL sample of 0.800 M NaOH was titrated with an acid. It required 25.0 mL of the acid to reach the equivalence point. The molarity of the acid is

Calculate the concentration of an HCl solution if 100.0 mL of the HCI required 33.00 mL of 0.2000 M Mg(OH)₂ to reach the titration endpoint. Mg(OH)2 + 2 HCl → 2 H₂O + MgCl₂ There is enough information to calculate the moles of base but not the moles of acid. mol b. 0.1320 i. OH-1 p. CO3² W. PO4³ a. 0.1000 h. CaCO3(s) o. H₂PO4-¹ v. H₂CO3 bb. 3.300 x 10-² hh. 2.16 x 10-6 L)(- mol Mg(OH)3)(- mol HCI c. HX j. H30+1 q. HCO3-¹ x. HCzH3Oz dd. 2 jj. 0.2000 d. Al+3 k. SO4² r. H₂S CC. 3 ii. 1.60 x 10-5 e. CO₂ 1. Mg+2 S. HS-1 y. C₂H30₂-1 ee. 1 = mol HCI mol Mg(OH)₂ M HCI f. CaF2(aq) m. C-1 t. S-2 z. 1 x 10-14 ff. 0.099536 kk. 6.600 x 10-³ g. HF(aq) n. HPO 2 u. H₂O(lig) mol Mg(OH)3 aa. 1.32 x 102 gg. 3.95 x 10-3 mol HCI