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Chapter 12: Q A E (page 282)

Potassium ion in a 250.0 (卤0.1) mL water sample was precipitated with sodium tetraphenylborate:

K++(C6H5)4B-鈫�KB(C6H5)4(s)

The precipitate was filtered, washed, dissolved in an organic solvent, and treated with excess Hg (EDTA)2-:

4HgY2-+(C6H5)4B-+4H2O鈫�H3BO3+4C6H5Hg++4HY3-+OH-

The liberated EDTA was titrated with 28.73 (卤0.03) mL of 0.043 7 (卤0.000 1) M Zn2+. Find [K+] (and its absolute uncertainty) in the original sample.

Short Answer

Expert verified

The amount of K+ in the original sample will be 1.256(卤0.003) mM

Step by step solution

01

Given information

Amount of water sample = 250 mL

For each mole of K+ involved in the first reaction , 4 moles of EDTA generated in the second reaction .

Amount of Zn2+ used for EDTA titration=28.73 (卤0.03) mL of 0.043 7 (卤0.000 1) M

02

EDTA

EDTA also known as Ethylenediaminetetraacetic acid is one type of chelating agent used in different industries, medications and laboratory applications. It is generally used to treat heavy metal toxicity, lead poisoning, coronary artery disease, neurotoxicity etc.

03

Determine K+

K+=14molesofZn2+Volumeofsample=1428.73卤0.03mL0.0437卤0.0001M250卤0.1=14卤0%28.73卤0.104%0.0437卤0.229%250卤0.040%=1.256卤0.265脳10-3M=1.256卤0.003mM

Therefore, the amount of K+ in the original sample will be 1.256(卤0.003) mM.

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Most popular questions from this chapter

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 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

Spreadsheet equation for auxiliary complexing agent. Consider the titration of metal M (initial concentration = CM, initial volume = VM) with EDTA (concentration = CEDTA, volume added = VEDTA) in the presence of an auxiliary complexing ligand (such as ammonia). Follow the derivation in Section 12-4 to show that the master equation for the titration is

蠒=CETDAVETDACMVM=1+Kf"[M]free-[M]free+Kf"[M]freeCMKf"[M]free+[M]free+Kf"[M]free2CETDA

where role="math"> is the conditional formation constant in the presence of auxiliary complexing agent at the fixed pH of the titration (Equation 12-18) and [M]free is the total concentration of metal not bound to EDTA. [M]free is the same as [M] in Equation 12-15. The result is equivalent to Equation 12-11, with [M] replaced by [M]free and Kf replaced by Kf".

Explain why the change from red to blue in Reaction 12-19 occurs suddenly at the equivalence point instead of gradually throughout the entire titration.

Spreadsheet equation for formation of the complexes ML and ML2.Consider the titration of metal M (initial concentration = CM, initial volume = VM) with ligand L (concentration = CL, volume added = VL), which can form 1:1 and 2 : 1 complexes:

M+L饾啅ML尾1=[ML][M][L]M+2L饾啅ML2尾2=[ML2][M][L]2

Let 伪M be the fraction of metal in the form M, 伪ML be the fraction in the form ML, and 伪ML2be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

role="math" localid="1667801924683" 伪M1=11+尾1[L]+尾2[L]2伪ML=尾1[L]1+尾1[L]+尾2[L]2伪ML2=尾2[L]21+尾1[L]+尾2[L]2

The concentrations of ML and ML2are

[ML]=伪MLCMVMVM+VL[ML2]=伪ML2CMVMVM+VL

because CMVMVM+VLis the total concentration of all metal in the solution. The mass balance for ligand is

[L]+[ML]+2[ML2]=CMVMVM+VL

By substituting expressions for [ML] and [ML2] into the mass balance, show that the master equation for a titration of metal by ligand is

蠒=CLVLVM+VM=伪ML+2伪ML2+LCM1-LCL
See all solutions

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