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27 Nov 2019
A 12.63-mL sample of aqueousNH3 was reacted with 50.00-mL of 0.1025 M HCl. Theexcess acid is then neutralized by titrating the sample to aphenolphthalein end-point with 25.98-mL of 0.1445 M NaOH. Thismethod is known as an indirect method of analysis.
a. Explain why the reaction between NaOH and HClallows you to determine the molarity of the NH3solution. b. Show the series of calculations required tocalculate the molarity of NH3 in the aqueoussolution.
A 12.63-mL sample of aqueousNH3 was reacted with 50.00-mL of 0.1025 M HCl. Theexcess acid is then neutralized by titrating the sample to aphenolphthalein end-point with 25.98-mL of 0.1445 M NaOH. Thismethod is known as an indirect method of analysis.
a. Explain why the reaction between NaOH and HClallows you to determine the molarity of the NH3solution.b. Show the series of calculations required tocalculate the molarity of NH3 in the aqueoussolution.
Collen VonLv2
1 May 2019