/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q32P A 50.0-mL sample containing Ni2+... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Q32P (page 285)

A 50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.050 0 M Zn2+. What was the concentration of Ni2+ in the original solution?

Short Answer

Expert verified

The concentration of Ni2+ in the original solution is 0.02 M

Step by step solution

01

Given Information

Amount of sample taken= 50.0 mL

Amount of EDTA required for titration =25 mL of 0.05M EDTA

Amount of Zn2+ required for back titration= 5 mL of 0.050 M Zn2+

02

Determine the number of moles of EDTA and Zn2+ required

Number of moles of EDTA required

=25mL0.050MEDTA=1.25mmol

Number of moles of Zn2+ required

=(5mL)(0.050EDTA)=0.25mmol

03

Determine the concentration of Ni2+

Let the number of mols of Ni2+ be x

mmolEDTA=mmolNi2++mmolZn2+1.250mmolEDTA=xmmolNi2++0.250mmolZn2+x=1mmolNi2+

50.0-mL of sample containing Ni2+ was taken

Therefore, concentration of Ni2+in the solution

=1mmol50mL=0.02M

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn2+ will be required for back titration?

According to Appendix I, Cu2+ forms two complexes with acetate:

Cu2++CH3CO2鈭�鈬�Cu(CH3CO2)+鈥夆赌夆赌夆赌夆赌夆赌夆赌�尾1(=K1)Cu2++2CH3CO2鈭�鈬�Cu(CH3CO2)2鈥夆赌夆赌夆赌夆赌夆赌夆赌�尾2

(a) Referring to Box 6-2, find K2 for the reaction

Cu(CH3CO2)++CH3CO2鈭�鈬�Cu(CH3CO2)2(aq)鈥夆赌夆赌�K2

(b) Consider 1.00 L of solution prepared by mixing 1.00 脳 10-4 mol Cu(ClO4)2 and 0.100 mol CH3CO2Na. Use Equation 12-16 to find the fraction of copper in the form Cu2+

Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn2+ at the following volumes of added EDTA and sketch the titration curve:

(a) 0 mL (b) 20.0 mL (c) 40.0 mL (d) 49.0 mL (e) 49.9 mL (f) 50.0 mL (g) 50.1 mL (h) 55.0 mL (i) 60.0 mL

Calculate [HY3-] in a solution prepared by mixing 10.00 mL of 0.010 0 M VOSO4, 9.90 mL of 0.010 0 M EDTA, and 10.0 mL of buffer with a pH of 4.00
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.