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Chapter 8: Q4P (page 183)

Calculate the ionic strength of (a) 0.008 7 M KOH and (b) 0.000 2 $M - La {({IO}_{3})}_{3}$ (assuming complete disassociation at this low concentration and no hydrolysis reaction to make ${LaOH}^{2 +}$ ).

Short Answer

Expert verified

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

0.000 2 $M - La {({IO}_{3})}_{3}$ is 0.0012M

Step by step solution

01

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

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

$渭 = \frac{1}{2} (c_{1} {z_{1}}^{2} + c_{2} {z_{2}}^{2} + . . .)$

, 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 be $KOH 謴 k^{+} + {OH}^{-}$

role="math" localid="1654835701912" $z_{1} = + 1 z_{2} = - 1 渭 = \frac{1}{2} (0.0087 脳 {(+ 1)}^{2} + 0.0087 脳 {(- 1)}^{2}) - 0.0087 M$

02

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

Next, for 0.000 2 $M - La {({IO}_{3})}_{3}$ , in this case both ions have different concentration.

$La {({IO}_{3})}_{3} + H_{3} 翱謴 La {(OH)}^{2} + 3 {IO}_{3} + H^{+} 渭 = \frac{1}{2} (0.0002 脳 {(+ 3)}^{2} + 0.0006 脳 {(- 1)}^{2}) = 0.0012 M$

Thus, the ionic strength for 0.008 7 M KOH is 0.0087M and for $0.000 2 M - La {({IO}_{3})}_{3}$ is 0.0012 M

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Most popular questions from this chapter

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Sodium acetate hydrolysis treated by Solver with activity coefficients.

(a) Following the NH₃ example in Section 8-5, write the equilibria and charge and mass balances needed to find the composition of 0.01 M sodium acetate (Na⁺A^-). Include activity coefficients where appropriate. The two reactions are hydrolysis (pK_b = 9.244) and ionization of H₂O.

(b) Including activity coefficients, set up a spreadsheet analogous to Figure 8-12 to find the concentrations of all species. Assign an initial value of ionic strength = 0.01. After the rest of the spreadsheet is set up, change the ionic strength from the numerical value 0.01 to the correct formula for ionic strength. This two-step process of beginning with a numerical value and then going to a formula is necessary because of circular references between ionic strength and concentrations that depend on ionic strength. There are four unknowns and two equilibria, so use Solver to find 4 - 2 = 2 concentrations (pC values). Solver does not find both pC values at the same time well in this problem. Execute one pass to find both pC values by varying pA and pOH to minimize $危 {b_{i}}^{2}$ . Then vary only pA to minimize $危 {b_{i}}^{2}$ . Then vary only pOH to minimize $危 {b_{i}}^{2}$ . Continue alternating to solve for one value at a time as long as $危 {b_{i}}^{2}$ continues to decrease. Find [A^-], [OH^-], [HA], and [H⁺]. Find the ionic strength, pH =-log([H⁺] $纬^{+}$ ) and the fraction of hydrolysis = [HA]/F.

Using activities, calculate the pH of a solution containing 0.010 M NaOH plus 0.012 0 M LiNO₃ . What would be the pH if you neglected activities?

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