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Zarrin [17]
3 years ago
12

A chemist must prepare of potassium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask

about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to significant digits
Chemistry
1 answer:
maksim [4K]3 years ago
3 0

The question is incomplete; the complete question is:

A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

Answer:

4.5g (to 2 significant digits)

Explanation:

Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

To find the pH we can use the equation pH + pOH = 14.

Firstly to find the pOH:

13.00 + pOH = 14

pOH = 1.00

To find the [OH-]

Since

pOH= -log[OH^-]

[OH^-] = antilog (-pOH)

[OH^-]= antilog (-1)

[OH^-] = 0.1 molL-1

Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

Also, we know that concentration = number of moles/volume

we have the concentration and the volume now so we can calculate the number of number of moles as follows:

The 800mL volume is the same as 0.8L

0.1 molL-1= number of moles/0.8L

0.08 moles = number of moles

now we can calculate the amount of solid KOH required

the molar mass of KOH = 39 + 16 +1 = 56 gmol-1

56 x 0.08 moles = 4.48g

So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.

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Answer:

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Explanation:

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Molarity is expressed as:

M=\frac{molessolute}{Lsolution}

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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.

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\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.

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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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As a result, formula of magnesium oxide is MgO and not MgO_{2}.

The student write the equation as Mg + O_2 \rightarrow MgO_2, it is not correct.

Therefore, given equation will be balanced as follows.

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Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.

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pH of 3.

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