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hammer [34]
3 years ago
15

A solution is prepared by adding 0.10 mol of iron(III) nitrate, Fe(NO3)3, to 1.00 L of water. Which statement about the solution

is correct?a. The solution is basicb. The solution is acidic.c. The solution is neutral.d. The value of K_a for the species in solution must be known before a prediction can be made. e. The value of K_b for the species in solution must be known before a prediction can be made.
Chemistry
1 answer:
MrMuchimi3 years ago
6 0

Answer:

Option b is correct.

Explanation:

When Fe(NO_3)_3 is dissolved in water, Fe(OH)_2 and HNO_3 are formed.

Fe(NO_3)_3\; + 3H_2O \leftrightharpoons Fe(OH)_2\; + HNO_3

As Fe(OH)_2 is an weak base and HNO_3 is a strong acid, the resulting solution will be acidic in nature.

if K_b value is known, then pH of the solution can also be determined which will support prediction.

However, acidic and basic nature of the solution can be predicted without K_b value also.

So, among the given option, (b) is correct.

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Answer:

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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]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

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^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             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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