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3 years ago

You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?

1 answer:

3 0

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:

pH = pKa + log ([base]/[acid]) ---> 1

Where,

[base] = concentration of C2H3O2 in molarity or moles

[acid] = concentration of HC2H3O2 in molarity or moles

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

Therefore using equation 1,
4.24 = 4.74 + log (1 / x)

log (1 / x) = - 0.5

1 / x = 0.6065

x = 1.65

The required ratio of C2H3O2 /HC2H3O2 is 1:1.65 or 3:5.

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2H₂ + O₂ → 2H₂O

mole ratio of O₂ : H₂O is 1 : 2

∴ if moles of O₂ = 16 mol

then moles of H₂O = (16 mol × 2)

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3 years ago

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The [α] of pure quinine, an antimalarial drug, is −165. If a solution contains 86% quinine and 14% of its enantiomer (ee = 72%),

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Answer : The [α] for the solution is, -118.8

Explanation :

Enantiomeric excess : It is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.

Mathematically,

$\%\text{ Enantiomer excess}=\%\text{ Major enantiomer}-\%\text{ Minor enantiomer}$

Given:

% major enantiomer = 86 %

% minor enantiomer = 14 %

Putting values in above equation, we get:

$\%\text{ Enantiomer excess}=86\%-14\%=72\%$

$\text{ Enantiomer excess}=\frac{72}{100}=0.72$

Now we have to calculate the [α] for the solution.

$[\alpha]=\text{Enantiomer excess}\times [\alpha]_{Pure}$

$[\alpha]=0.72\times -165$

$[\alpha]=-118.8$

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3 years ago

When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)

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Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ = 0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂ + 2K₃PO₄ → Ba₃(PO₄)₂ + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

Ba(NO₃)₂ : Ba₃(PO₄)₂

3 : 1

0.182 × 10⁻³ : 1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

K₃PO₄ : Ba₃(PO₄)₂

2 : 1

2.537 × 10⁻³ : 1/2 × 2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by Ba(NO₃)₂ are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

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