Which of the following statements is TRUE about precipitation titration? Choose the best answer. It is based on reactions that yield ionic compounds of limited solubility or sparingly soluble substances It forms a deep blood-red colored complex at endpoint. It uses silver chloride as the standard titrant. It requires the use of a secondary titrant to back titrate the excess amount of titrant used to react with the analyte.

Which of the following statements is TRUE about precipitation titration? Choose the best answer. It is based on reactions that yield ionic compounds of limited solubility or sparingly soluble substances It forms a deep blood-red colored complex at endpoint. It uses silver chloride as the standard titrant. It requires the use of a secondary titrant to back titrate the excess amount of titrant used to react with the analyte.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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We're using phenolphthalein as the indicator, the end point is reached when one drop of the NaOH solution turns the entire solution from colorless to pink that persists for at least 30 min. True False
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What is a primary reason for needing to standardize strong base titrants like KOH? Select the single best answer. OKOH is not very soluble in aqueous solutions, thus making the concentrations difficult to determine. KOH reacts with moisture in the air, thereby making accurately-determined concentrations of their solutions difficult. O Standardization is the process of reacting a strong base, like KOH, with an acid. O Standardization is the process of dissolving KOH in aqueous solution. OKOH is a very inexpensive, highly soluble, and easy-to-use base.
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20 ml of a 100 mg/l phenol solution in water is placed in a vial with 20 ml of a fluorocarbon solvent. The organic phase (fluorocarbon solvent) is immiscible with water. The vial is completely filled with the two phases, thus there is no air or contaminant volatilization. After an hour of agitation, the two phase system reaches equilibrium. The two phases are separated and the fluorocarbon is analyzed for total phenol. The resulting phenol concentration in the organic phase is 16.0 mg/l. Compute the unitless partition coefficient (KF-W) for phenol. Compare your calculation with KF-W obtained from an empirical relationship: log10KF-W 0.71 - 0.862log10 CWsat. KF-W means the partition between the fluorocarbon solvent, F, and water, W. Only typed solution.
A 150 mL solution of NaOH is titrated with HI. Phenol red is the indicator, which is red at high pH values and yellow at low pH values. The indicator be [ Select ] when you first put it into your analyte, before any titrant is added. The indicator will be [ Select] at the endpoint. The indicator will be [Select] V if you overshoot the endpoint and add too much acid.
A student does not filter his/her saturated solution before titrating. Will the calculated Ksp probably be too high, too low, or unaffected? Why? Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected. The remaining solids would absorb excess H30+ ions during the titration, making the calculated value of [OH ], and thus Ksp, too low. O The remaining solids would dissolve during the titration to give more OH¯. The calculated Ksp would therefore be too high.