A. If you were to stop the titration and add 2.0 mL of distilled water to the an

Question

A. If you were to stop the titration and add 2.0 mL of distilled water to the analyte flask, which of the the following would change following the addition of the water?

The volume of the analyte solution.

The moles of H3O+.

The moles of OH-.

The pH of the analyte solution.

The volume of OH- needed to reach the endpoint.

The concentration of the H3O+ in the analyte solution.

B. Select all of the dominant species (present in largest amounts) that would exist in solution when you are halfway to the endpoint of this titration.

HCl, H3O+, OH-, Cl-, NaOH, Na+

       C. You are titrating a 11.2 mL sample of unknown Ca(OH)2 solution with a known solution of HNO3. If it took 19.4 mL of a 1.20 M solution of HNO3, what is the concentration of the original Ca(OH)2 solution? Hint: watch your stoichiometry.

         D. Select all of the molecules or ions that will be the dominant species in solution at the exact equivalence point of this titration. Do not include the solvent or ions that are there only because of the autoionization of water.

H3O+, Ca2+, Ca(OH)2, OH-, HNO3, NO3-

Explanation / Answer

From the statement i understand that the analyte is an acid being titrated with NaOH so

if you add water to the analyte you will

increase the volume of the analyte solution

PH of the analyte will change (this is because ph comes from the concentration of H+ if you add more water the concentration of H will decrease and the ph will increase)

The concentration of H3O+ (H+) in the solution

B) halfway equivalence point means that you havent titrated all the acid , but you have already formed some NaCl salt

You wont have HCl , you have H and Cl

You have Na ions and Cl ions

so you have H3O+ , Cl- , Na+

c)

Ca(OH)2 + 2 HNO3 ==== 2H2O + Ca(NO3)2

there are 19.4 ml of 1.2 M of HNO3, get the number of moles

0.0194 * 1.2 = 0.02328 moles

1 mole of Ca(OH)2 needs 2 moles of HNO3 so, if you used 0.02328 moles of HNO3 then you have

0.02328 / 2 = 0.01164 moles of base

concentration of base is moles / volume

0.01164 / 0.0112 = 1.039M

D) at the exact equivalent point according to the reaction

Ca(OH)2 + 2 HNO3 ==== 2H2O + Ca(NO3)2

all the acid will be neutralized and there will be no base only the salt

Ca(NO3)2 this dissociates in Ca+2 and NO3-1

you will only have those ions in the solution

*if this answer is helpful dont forget to rate it =)

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