Calculate the ionic strength of (a) 0.008 7 M KOH and (b) 0.000 2 M-La(IO3)3(assuming complete disassociation at this low concentration and no hydrolysis reaction to makeLaOH2+ ).

Short Answer

Expert verified

The ionic strength for 0.008 7 M KOH is 0.0087M and for

0.000 2 M-La(IO3)3is 0.0012M

Step by step solution

01

Step 1:Finding the ionic strength of  0.008  7   M   KOH

The ionic strength μaa concentration of ions in a solution is calculated by using the following formula,

μ=12(c1z12+c2z22+...)

, wherestands for concentration and z stands for charge.

First the ionic strength of 0.008 7 M KOH can be calculated.

The reaction of disassociation would beKOH֏k++OH-

role="math" localid="1654835701912" z1=+1z2=-1μ=120.0087×+12+0.0087×-12-0.0087M

02

Finding the ionic strength of 0.000   2 M-    La(IO3)3

Next, for 0.000 2 M-LaIO33, in this case both ions have different concentration.

LaIO33+H3LaOH2+3IO3+H+μ=120.0002×+32+0.0006×-12=0.0012M

Thus, the ionic strength for 0.008 7 M KOH is 0.0087M and for 0.0002M-LaIO33is 0.0012 M

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