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2 years ago

Match the following.

Chemistry

1 answer:

Andru [333]2 years ago

5 0

Answer: Please see below for answers

Explanation: Matching appropriate labels , we have

1)3/4 of the way to second equivalence point of a diprotic acid/strong base titration-- pH=pka₂

equivalence point of a weak base/strong acid titration=pH<7

equivalence point of a strong acid/strong base titration= pH=7

equivalence point of a weak acid/strong base titration=pH>7

half-way to equivalence point of a weak acid/strong base titration pH =pka

where

pH gives the measure of the amount of concentration of hydrogen ions in an aqueous solution.

pKa is known as acid dissociation constant which explains the equilibrum at which a chemical species can give out or receive proton

pka₂ is the acid dissociation constant for the second ionization energy.

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A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

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Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

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$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

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1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

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