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3 years ago

4. Which of the following aromatic compounds undergoes nitration faster than benzene does? Explain your answer

Chemistry

1 answer:

Angelina_Jolie [31]3 years ago

4 0

Answer:

Ph-OH

Explanation:

Nitration of Aromatic compounds is an example for Aromatic Electrophilic substitution and its rate is enhanced by Electron donating group like OMe, OH, alkyl group and halides (least activating group) while reactivity is diminished by electron withdrawing like CN,NO2 and Ketone etc

Hence phenol is more reactive than benzene towards nitration.

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If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

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Answer:

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Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

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(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

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2 years ago

A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added

Ulleksa [173]

Answer: The pH change of the buffer is 0.30

Explanation:

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})$ .....(1)

We are given:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.306M$

$[C_6H_5NH_2]=0.418M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{initial}=14-8.99=5.01$

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-$

Initial: 0.418 0.124 0.306

Final: 0.294 - 0.430

Calculating the pOH by using using equation 1:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.430M$

$[C_6H_5NH_2]=0.294M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{final}=14-9.29=4.71$

Calculating the pH change of the solution:

$\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30$

Hence, the pH change of the buffer is 0.30

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