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2 years ago

What fraction of a 0.5 m acetic acid solution will be in the protonated form at ph 5.1? The pka of acetic acid is 4.8?

Chemistry

1 answer:

IRINA_888 [86]2 years ago

3 0

The acetic acid will dissociate as

We will use Hendersen Hassalbalch's equation here

pH = pKa + log [salt] / [acid]

Where

[Acid] = concentration of protonated form

5.1 = 4.8 + log [salt] / [acid]

1.99= [salt] / [acid]

[salt] +[acid] = 0.5

1.99 [acid] + [acid] = 0.5

[acid] = 0.167 M

[acid] / total concentration = 0.167 / 0.5 = 0.33.4

So percentage = 33.4 % or approx 33 %

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3 years ago

An object becomes electrically charged when:

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3 years ago

The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

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Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

- For the $BaCl_{2}$

M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

- For the $Na_{2}SO_{4}$

M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

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3 years ago

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

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Answer: Mass of $CO_2$ produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants = Mass of $CaCO_3$ + mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$ produced in this reaction was 6.56 grams

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