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Chapter 8: QBE (page 182)

Find the activity (not the activity coefficient) of the ${(C_{3} H_{7})}_{4} N^{+}$ (tetra propylammonium) ion in a solution containing ${(C_{3} H_{7})}_{4} N^{+} {Br}^{-} plus 0.0050 M {({CH}_{3})}_{4} N^{+} {CI}^{-}$

Short Answer

Expert verified

The activity of ${(C_{3} H_{7})}_{4} 鈥� N^{+}$ ion is 0.0046.

Step by step solution

01

Define the formula for activity species.

Any species' activity value is calculated by multiplying its concentration by its activity coefficient.

For a species C, the activity is given by

$A_{c} = [C] 纬_{c}$

where,

A - Activity

[c] - Concentration

- Activity coefficient.

02

Calculate the activity species.

Given

$0.0050 M (C_{3} H_{7}) N^{+} {Br}^{-}$ plus $0.0050 M {({CH}_{3})}_{4} N^{+} {CI}^{-}$

The ionic strength of this solution is $渭 = 0.010 M$

The size of the ion is ${(C_{3} H_{7})}_{4} 鈥� N^{+}$ is 800pm

$渭 = 0.010 M,$

At $纬 = 0.912$ for an ion of charge ${}_{-}^{+}1$ with $伪 = 800 pm$

The activity ofrole="math" localid="1654762931695" ${(C_{3} H_{7})}_{4} N^{-}$ by using activity coefficient and concentration values

$A_{{(C_{3} H_{7})}_{4}} N^{+} = [{(C_{3} H_{7})}_{4} N^{+}] 纬_{{(C_{3} H_{7})}_{4} N^{-}} = (0.0050) (0.912) = 0.0046$

The activity of ${(C_{3} H_{7})}_{4} N^{+}$ ion is 0.0046.

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Most popular questions from this chapter

Interpolate in Table 8-1 to find the activity coefficient of $H^{+}$ when $渭 = 0.030 M$

Ion pairing. As in Problem $8 - 30$ , find the concentration, ionic strength, and ion pair fraction in localid="1654942556135" $0.025 F {Na}_{2} {SO}_{4}$ . Assume that the size of the ${NaSO}^{-}_{4} = 500 pm$

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Including activity coefficient, find the concentration of ${Ba}^{2 +}$ in $0.100 M {({CH}_{3})}_{4} {NIO}_{3}$ solution saturated with $Ba {({IO}_{3})}_{2}$ .

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