A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wishes to determine its exact concentration. A 25.00 mL portion of the HCI solution is transferred to a flask, and after a few drops of indicator are added, the HCI solution is titrated with 0.07575 M NAOH solution. The titration requires exactly 38.92 mL of the standard NaOH solution to reach the end point. What is the molarity of the HCL solution? It takes 46

Fundamentals Of Analytical Chemistry
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Chapter16: Applications Of Neutralization Titrations
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is this the correct answer for 3a?

a. A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wishes to
determine its exact concentration. A 25.00 mL portion of the HCI solution is transferred to a
flask, and after a few drops of indicator are added, the HCI solution is titrated with 0.07575 M
NaOH solution. The titration requires exactly 38.92 mL of the standard NaOH solution to reach
the end point. What is the molarity of the HCL solution?
b. It takes 46.22 mL of a 1.021 M NaOH solution to neutralize a solution of 4.4567 g of an
unknown monoprotic acid in 80.00 mL of water. Calculate the molecular weight of the acid.
4(D) - 5
2.
NaOt
aq)
HClca)
30)
5,00mL
mol/
Transcribed Image Text:a. A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wishes to determine its exact concentration. A 25.00 mL portion of the HCI solution is transferred to a flask, and after a few drops of indicator are added, the HCI solution is titrated with 0.07575 M NaOH solution. The titration requires exactly 38.92 mL of the standard NaOH solution to reach the end point. What is the molarity of the HCL solution? b. It takes 46.22 mL of a 1.021 M NaOH solution to neutralize a solution of 4.4567 g of an unknown monoprotic acid in 80.00 mL of water. Calculate the molecular weight of the acid. 4(D) - 5 2. NaOt aq) HClca) 30) 5,00mL mol/
Na Cliaz)
0.lmol/L
HF 25.00ml
MANOH=D0.07575 mol/L
VNGOH= 38.92mL
%3D
/NO
NT
3(0.07575moll003892:
100029L18, mol.
TAN
Oxn-
0.002a48 mol
tmol NaO4
0.00948, mol
Transcribed Image Text:Na Cliaz) 0.lmol/L HF 25.00ml MANOH=D0.07575 mol/L VNGOH= 38.92mL %3D /NO NT 3(0.07575moll003892: 100029L18, mol. TAN Oxn- 0.002a48 mol tmol NaO4 0.00948, mol
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