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18 Dec 2019

It took 32.15 mL of 0.0953 M NaOH to reach the equivalence point of a titration when 25.00 mL of unknown concentration of HCl was titrated. Determine the molarity of the HCl solution.

_HCl(aq) + _NaOH ï _NaCl(aq) + _H2O(l)

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You add 12.4 mL of 0.18 M Ba(OH)2 to 25.00 mL of HCl to reach the equivalence point of the titration. Determine the concentration of HCl with the reaction shown below.

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A titration of 25.00 mL of a solution of the weak base aniline, C₆H₅NH₂ requires 25.67 mL of 0.1750 M HCl to reach the equivalence point.

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periwinklechimpanzee385

1. Determine the volume of 1M NaOH that is required to reach the equivalence point with 25.00 mL of HCl (of unknown concentration). From there, calculate the original concentration of the unknown HCl solution.

Volume of 1M NaOH used = 2.80 mL

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It took 32.15 mL of 0.0953 M NaOH to reach the equivalence point of a titration when 25.00 mL of unknown concentration of HCl was titrated. Determine the molarity of the HCl solution.

_HCl(aq) + _NaOH ï _NaCl(aq) + _H2O(l)

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It took 32.15 mL of 0.0953 M NaOH to reach the equivalence point of a titration when 25.00 mL of unknown concentration of HCl was titrated. Determine the molarity of the HCl solution. _HCl(aq) + _NaOH ï _NaCl(aq) + _H2O(l)

Related textbook solutions

Basic Chemistry

Principles of Chemistry Molecular Approach

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It took 32.15 mL of 0.0953 M NaOH to reach the equivalence point of a titration when 25.00 mL of unknown concentration of HCl was titrated. Determine the molarity of the HCl solution.

_HCl(aq) + _NaOH ï _NaCl(aq) + _H2O(l)