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Salsk061 [2.6K]
3 years ago
13

Formulate the sequence of steps needed to prepare 250.0 mL of a 0.005 M aqueous solution from solid KMnO4 and pure water.

Chemistry
1 answer:
S_A_V [24]3 years ago
3 0

Answer:

Molarity is a unit that measures how much moles of solute dissolved in a liter of solvent. Molarity expressed using capital M while molarity, a different unit, expressed using lower case m.  

We want to make 0.005 M solution which means we need 0.005 moles of KmnO4 per liter of water. First, we have to calculate how many grams of KMnO4 we need for the solution.

We want to make 250ml solution, so the number of moles of KMnO4 we need will be:  0.005 mol/liter *(250 ml * 1liter/1000ml)= 0.005 mol/liter  * 1/4 liter = 0.00125 moles

The molecular mass of KMnO4 is 158g/mol, so the mass of KMnO4 we need will be:  0.00125 moles *  158g/mol=  0.1975 grams

We know that we need 0.1975 g of KMnO4, now we weigh them and put it inside a dish. After that, we prepare Erlenmeyer or a volumetric flask filled with water half of the volume needed(125ml). Pour the weighted solute into the flask, stir until all solute dissolved.

Then we add water to the container slowly until its volume reaches the 250ml mark.

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C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED']  = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

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E + D ⇄ ED → K_a = [ED] / [D][E]   (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED]  (dissociation constant)

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[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

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[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

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ΔG_1 = -8184.633  * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED']                                  E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED']  = 0.0579

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