A 1.986 g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt ( ACES − K + , MW = 220.29 g/mol) is dissolved in 73.73 mL of water.To the solution, 14.73 mL of HCl is added, resulting in a pH of 6.56 . Calculate the concentration of the HCl solution. The p K a of ACES is 6.85.
A 1.986 g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt ( ACES − K + , MW...
Question 20 of 32 > A 1.069 g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES K'. MW = 220.29 g/mol) is dissolved in 87.54 mL of water. To the solution, 11.62 mL of HC is added, resulting in a pH of 7.42. Calculate the concentration of the HCl solution. The pk, of ACES is 6.85. [HC] = MacBook Pro 8 90
1.800 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 89.94 mL of water. 29.91 mL of HCl is added to the solution, resulting in a pH of 7.24. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.
A 1.944 g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES-K+, MW = 220.29 g/mol) is dissolved in 52.38 mL of water. To the solution, 27.66 mL of HCl is added, resulting in a pH of 6.95. Calculate the concentration of the HCl solution. The pK, of ACES is 6.85. (HCI) = 3.80 x10-6
A 1.982g sample of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt ( ACES−K+, MW = 220.29 g/mol) is dissolved in 87.51 mL of water.To the solution, 21.98 mL of HCl is added, resulting in a pH of 7.27. Calculate the concentration of the HCl solution. The p?a of ACES is 6.85. I have calculated this several times and have gotten the same answer, 0.112754 M. However, Sapling says that I am wrong. Then Sapling gives these instructions: To begin, calculate the original number...
7. Lactic acid (CSH COOH) is a weak monoprotic acid whose K, value is 1.4x10". Its salt potassium lactate is added to food products as a preservative and to inhibit growth of bacteria. What is the pH of a solution whose molar concentration of potassium lactate (CSHCOO-K) is 0.290 M? Caution! This problem does not say that lactic acid is dissolved in water, it is potassium lactate that is added to the water.
7. A 0.653 g of a monoprotic solid acid (mw= 157 g/mol) is dissolved in water to produce a 50.0 mL solution. The pH of this solution is measure to be pH = 2.13. Determine the ionization constant (Ka) of the acid. 8. A buffer is 0.50 M CH3COOH(aq) and 0.50 M Na(C6H-C00) (aq). For CH2COOH pKa = 4.20 a) Calculate the pH of the buffer solution. b) Calculate the pH after the addition of 40 mL of a 0.150...
A 0.1358-g portion of potassium iodate (MW 214.00), about 2 g of potassium iodide, and 2 mL of 6 M hydrochloric acid were dissolved in 25 mL of distilled water. The triiodide formed during the ensuring reaction was titrated to the starch endpoint with 31.94 mL of a thiosulfate solution. A 25.00-mL triiodide sample solution was titrated to the endpoint with 21.33 mL of the standardized thiosulfate solution. Calculate the concentrations of the thiosulfate solution and the triiodide solution.
A 100.0 mL100.0 mL solution containing 0.927 g0.927 g of maleic acid (MW=116.072 g/mol)(MW=116.072 g/mol) is titrated with 0.275 M KOH.0.275 M KOH. Calculate the pH of the solution after the addition of 58.0 mL58.0 mL of the KOHKOH solution. Maleic acid has p?apKa values of 1.92 and 6.27. pH= At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as H2M,H2M, HM−,HM−, and M2−,M2−,...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
A 1.20 gram sample of an unknown monoprotic acid is dissolved in 25.0 mL of water and titrated with a a 0.485 M aqucous potassium hydroxide solution. It is observed that after 7.33 milliliters of potassium hydroxide have been added, the pH is 4.515 and that an additional 13.3 mL of the potassium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? (2) What is the value of K, for the...