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Shalnov [3]
3 years ago
13

So my chem experiment asks us: What is the molarity of a 6.0 ppm solution (it's a 6.0 ppm chlorophyll solution, hexane is the so

lvent)
Now, I've found a few ways to solve for it but they both give me different values. If it's possible, could you just tell me which is right.
#1 Way: 6parts/1million --> 0.006g/L --> take 0.006g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 *10^(-6) moles
put 6.71524 *10^(-6) over 1L --> 6.71524 *10^(-6) M
0R 6parts/1million --> 6g/1000,000g --> take 6g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 *10^(-3) moles
use hexane's density to convert 1000,000g to (mL first then) L (density: 1mL/0.6548g) --> 1527.18388 L
put 6.71524 *10^(-3) moles over 1527.18388 L --> 4.397*10^(-6) M
OR 6parts/1million -->6g/1000,000g take 6g convert w/ molar mass chlorophyll (1 mole/ 893.49 g) --> 6.71524 *10^(-3) moles
put 6.71524 *10^(-3) moles over 1000,000 --> 6.71524 *10^(-9) M
I think the second one is the right one but I'm not sure. Could you please help me? Thanks!
Chemistry
2 answers:
Alex73 [517]3 years ago
8 0
The second approach is correct. The other two approaches are not correct because they are incomplete; first approach would have been right IF the 6.0 ppm was MEASURED in hexane. Third approach cannot be right since it calculates moles and grams but not L. 


Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

The first option is correct.

Explanation:

Concentration of chlorophyll is the ppm = 6.0 ppm

(1 ppm = 1 mg/kg)

6.0 ppm = 6.0 mg/kg

Mass of chlorophyll = 6.0 mg= 0.006 g (1 mg = 0.001 g)

Moles of chlorophyll =\frac{0.006 g}{893.49 g/mol}=6.71524\times 10^{-6} mol

volume of the solution = 1 L

Concentration(c) = \frac{n}{V(L)}

Where : n = moles of the compound

V = Volume of the solution

Concentration of the chlorophyll :

\frac{6.71524\times 10^{-6} mol}{1 L}=6.71524\times 10^{-6} M

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The pH of this solution is 1,350

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The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺        Kₐ₃ = 4,20x10⁻¹⁰  (1)

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The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

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The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄   ⇄      H₂PO4⁻         +        H⁺           Kₐ₁ = 7,50x10⁻³   (3)

X mol     ⇄  (1x10⁻³-X) mol  + (1x10⁻³-X) mol                            (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

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