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Both AgCl and AgI dissolve in $\mathrm{NH}_{3}$ …
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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115

Problem 103 Hard Difficulty

Calculate the concentration of $\mathrm{Cd}^{2+}$ resulting from the dissolution of $\mathrm{CdCO}_{3}$ in a solution that is 0.250 $\mathrm{M}$ in $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, 0.375 \mathrm{M}$ in $\mathrm{NaCH}_{3} \mathrm{CO}_{2},$ and 0.010 $\mathrm{M}$ in $\mathrm{H}_{2} \mathrm{CO}_{3}$ .

Answer

$1 \times 10^{-5} \mathrm{M}$

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Chemistry 102
Chemistry

Chapter 15

Equilibria of Other Reaction Classes

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Chemical Equilibrium

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Video Transcript

The expression for dissociation or solubility of c d, carmina box, late c d, 2, positive plus c, o 3 to negative and k value is equal to 2.510 raised to power negative 16 point. We substitute the concentration into k, s p formula and calculated as so concentration h, 3 o positive equals k a h, 4, a c divided by negative substituting the .375 now moving towards k. A of h, 2 c 3 gives h 3, o positive and co 3 negative divides by h, 2 c 3. H, 3, o positive substituting the values 3.5810 raised to power negative 4 pint from k, a and c o 3 to negative concentration. We can determine the value of k. A so ka will be equal to concentration for c o 3 to negative, divided by 3.5810, raise to the power negative 4 rearranging the equation: cero 3 to negative terms to be 3.5810 or negative 4 divided. By turns out to be 2.0910 raised to the power negative 9, the concentration for cadmium can be calculated using as formula so c. D 2 positive will be equal to 2.09 multiply 10 raised to 1 negative 9 and camumbine concentration terms out to be.

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

Chemistry

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