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In three successive titrations of exactly 25 mL aliquots of 0.1010M hydrochloric acid against a solution of sodium hydroxide, thedelivered volumes of hydroxide were 26.00 mL, 26.50 mL, and 26.46mL.

a) what mean volume of NaOH should be used to calculate themolarity of the acid?

b) calculate the molarity of the base.

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Elin Hessel
Elin HesselLv2
28 Sep 2019

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Related questions

Hello Chegg family i have a few titration and oxidation problems. Thank You!

1.) What volume of a 0.367 M hydrochloric acid solution is required to neutralize 14.3 mL of a 0.177 Msodium hydroxide solution?
??mL hydrochloric acid

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??mL barium hydroxide

2.) An aqueous solution of hydrochloric acid is standardized by titration with a 0.177 M solution of calcium hydroxide.
If 14.3 mL of base are required to neutralize 27.8 mL of the acid, what is the molarity of the hydrochloric acid solution?
??M hydrochloric acid

An aqueous solution of barium hydroxide is standardized by titration with a 0.142 M solution of perchloric acid.
If 24.1 mL of base are required to neutralize 13.6 mL of the acid, what is the molarity of the barium hydroxide solution?
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3.) Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as aprimary standard. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP.
If 37.5 mL of a sodium hydroxide solution are needed to neutralize 0.957 grams of KHP, what is the molarity of the sodium hydroxide solution?
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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 = MUsing 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 = molThis 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 = molMoles of HCl required to neutralize the ammonia = molThis 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 = gMass of Ni(NO3)2 in the sample = gDivide 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 = molMole ratio of ammonia to Ni(NO3)2 in the complex =

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 = MUsing 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 = molThis 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 = molMoles of HCl required to neutralize the ammonia = molThis 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 = gMass of Ni(NO3)2 in the sample = gDivide 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 = molMole ratio of ammonia to Ni(NO3)2 in the complex =

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