MasteringChemistry Homework 2-CH10.8-11, 12.1-5-Google Chrome secure https://ses

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MasteringChemistry Homework 2-CH10.8-11, 12.1-5-Google Chrome secure https://session.masteringchemistry.com myct/itemView?assignmentProblemID=97541685 Homework 2-CH10.8-11, 12.1.5 Animation - Acid-Base Titration 12 of 32 Part B A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction fask. Classify the following conditions based orn whether they are before the equivalence point, at the equivalence point, or after the equivalence point. Drag the appropriate items to their respective bins. View Available Hint(s) Reset Help 200 mL of 1.00 MNaCH 100 mL of 1.00 M NaOH 50.0 nL of 1.00 M NaOH 150 mL of 1.00 W NaOH10.0 mL of 1.00 MNaCH 5.00 mL of 1.00 M NaOHH Before equivalence point At equivalence point After equivalence point 6:11 PM Type here to search 1/23/2018 1

Explanation / Answer

Ans. Balanced equation:      HCl(aq) + NaOH(aq) ----------> NaCl(aq) + H2O

According to the stoichiometry of balanced reaction, 1 mol HCL is neutralized by 1 mol NaOH.

# Volume of HCl solution = 100 mL = 0.100 L                            ; [1 L = 1000 mL]

Moles of HCl to be neutralized = Molarity of HCl x Volume of solution in liters

                                                = 1.00 M x 0.100 L

                                                = 0.100 moles

At equivalence point, the moles of NaOH added must be equal to the moles of HCl being titrated (or being neutralized).

So, if moles of NaOH added is less than 0.100 moles, the NaOH solution belongs to “before equivalence point” because equivalent point would not be reached.

If moles of NaOH added is greater than 0.100 moles, the NaOH solution belongs to “after equivalence point” because equivalent point would be surpassed by it.

# Calculate moles of NaOH in-

I. 200 mL of 1.00 M NaOH = 1.0 M x 0.200 L = 0.200 mol

II. 100 mL of 1.00 M NaOH = 1.0 M x 0.100 L = 0.100 mol

III. 50 mL of 1.00 M NaOH = 1.0 M x 0.050 L = 0.050 mol

IV. 150 mL of 1.00 M NaOH = 1.0 M x 0.150 L = 0.150 mol

V. 10 mL of 1.00 M NaOH = 1.0 M x 0.010 L = 0.010 mol

VI. 5 mL of 1.00 M NaOH = 1.0 M x 0.005 L = 0.005 mol

# Conclusion:

Before equivalence point: 5 mL of 1.00 M NaOH, 10 mL of 1.00 M NaOH, 50 mL of 1.00 M NaOH

At equivalence point: 100 mL of 1.00 M NaOH

After equivalence point: 200 mL of 1.00 M NaOH, 150 mL of 1.00 M NaOH

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