The concentration of a hydrochloric acid solution is approximately 0.25 M but th

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The concentration of a hydrochloric acid solution is approximately 0.25 M but the solution must be standardized by titration against sodium hydroxide to determine the precise concentration. If the molarity of the sodium hydroxide solution is 0.2053 M and the volume required to reach the end point in the titration against 25.00 mL of the hydrochloric acid is 29.57 mL, what is the molarity of the hydrochloric acid solution? Molarity of HCl = M Using an analytical balance, 0.1682 grams of the nickel(II) ammine complex was weighed by difference and placed in an Erlenmeyer flask. Exactly 25 mL of the standardized hydrochloric acid solution was added to neutralize the ammonia in the complex. The excess hydrochloric acid remaining in the Erlenmeyer flask was titrated against the sodium hydroxide. 14.55 mL of sodium hydroxide was required to reach the end point. How many moles of sodium hydroxide were required? Moles of NaOH required = mol This equals the number of moles of excess HCl remaining in the flask. How many moles of HCl are there in 25.00 mL of the hydrochloric acid solution, and therefore how many moles of HCl were required to neutralize the ammonia in the nickel(II) ammine complex? Moles of HCl in 25.00 mL = mol Moles of HCl required to neutralize the ammonia = mol This equals the number of moles of ammonia in your sample of the complex. Multiply by 17.031 to obtain the mass of ammonia. Subtract this from the mass of the complex (0.1682 grams) to obtain the mass of nickel ions and nitrate ions in the sample. Mass of ammonia in the sample = g Mass of Ni(NO3)2 in the sample = g Divide this mass by 182.70 to obtain the number of moles of Ni(NO3)2 in the sample and then calculate the mole ratio of Ni(NO3)2 to ammonia in the complex. Moles of Ni(NO3)2 in the sample = mol Mole ratio of ammonia to Ni(NO3)2 in the complex =

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The concentration of a hydrochloric acid solution is approximately 0.25 M but the solution must be standardized by titration against sodium hydroxide to determine the precise concentration. If the molarity of the sodium hydroxide solution is 0.2000 M and the volume required to reach the end point in the titration against 25.00 mL of the hydrochloric acid is 31.76 mL, what is the molarity of the hydrochloric acid solution?

Molarity of HCl = 0.2541 M

Using an analytical balance, 0.1601 grams of the nickel(II) ammine complex was weighed by difference and placed in an Erlenmeyer flask. Exactly 25 mL of the standardized hydrochloric acid solution was added to neutralize the ammonia in the complex. The excess hydrochloric acid remaining in the Erlenmeyer flask was titrated against the sodium hydroxide. 15.66 mL of sodium hydroxide was required to reach the end point.

How many moles of sodium hydroxide were required?

Moles of NaOH required = 3.132E-3 mol

This equals the number of moles of excess HCl remaining in the flask. How many moles of HCl are there in 25.00 mL of the hydrochloric acid solution, and therefore how many moles of HCl were required to neutralize the ammonia in the nickel(II) ammine complex?

Moles of HCl in 25.00 mL = 6.352E-3 mol

Moles of HCl required to neutralize the ammonia = 3.221E-3 mol

This equals the number of moles of ammonia in your sample of the complex. Multiply by 17.031 to obtain the mass of ammonia. Subtract this from the mass of the complex (0.1601 grams) to obtain the mass of nickel ions and nitrate ions in the sample.

Mass of ammonia in the sample = 5.485E-2 g

Mass of Ni(NO3)2 in the sample = 0.1052 g

Divide this mass by 182.70 to obtain the number of moles of Ni(NO3)2 in the sample and then calculate the mole ratio of Ni(NO3)2 to ammonia in the complex.

Moles of Ni(NO3)2 in the sample = 5.761E-4 mol

Mole ratio of ammonia to Ni(NO3)2 in the complex = 5.59

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