A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.

The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.

Explanation:

Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.  

The mass of the precipitate given is 22.73 grams.  

Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams

Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.  

And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.

Now the equation will become,  

3x/78 + 45.46-2x/92.94 = 0.647 moles

0.03846 x + 0.489 - 0.02152 x = 0.647  

0.01694 x + 0.489 = 0.647

0.01694 x = 0.158

x = 0.158/0.01694

x = 9.327 grams

Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,  

= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams

Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams

the mass of Co²⁺ will be,  

= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

b. examine and describe cells