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Chapter 12: Q30P (page 285)

What is meant by water hardness? Explain the difference between temporary and permanent hardness.

Short Answer

Expert verified

Water hardness is the measure of total concentration of Ca2+, Mg2+ or Fe2+ (basically cations) in water. The main difference between temporary and permanent hardness is the removal technique. Temporary hardness of water can be removed by boiling water whereas permanent hardness cannot be removed by boiling.

Step by step solution

01

Water Hardness

Water hardness is the measure of total concentration of Ca2+, Mg2+ or Fe2+ (basically cations) in water.

02

Permanent and Temporary Hardness

The carbonates and bicarbonates salts of calcium and magnesium causes temporary hardness in water. Permanent hardness in water is caused by the presence of sulfates and chlorides of calcium and magnesium.

03

Difference between Permanent and Temporary Hardness

The main difference is the removal technique. Temporary hardness of water can be removed by boiling water whereas permanent hardness cannot be removed by boiling.

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Most popular questions from this chapter

Find 伪Zn2+if free, unprotonated [NH3] = 0.02 M.

Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni2+ solution (containing excess Ni2+) to convert the cyanide into tetracyano nickelate (II):

4CN-+Ni2+鈫�Ni(CN)42-

Excess Ni2+ was then titrated with 10.15 mL of 0.01307 M EDTA. Ni(CN)42-does not react with EDTA. If 39.35 mL of EDTA were required to react with 30.10 mL of the original Ni2+ solution, calculate the molarity of CN- in the 12.73-mL sample.

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

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

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
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