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sveticcg [70]
3 years ago
6

A buffered solution containing dissolved aniline, C 6 H 5 NH 2 , and aniline hydrochloride, C 6 H 5 NH 3 Cl , has a pH of 5.47 .

A. Determine the concentration of C 6 H 5 NH + 3 in the solution if the concentration of C 6 H 5 NH 2 is 0.215 M. The p K b of aniline is 9.13.
Chemistry
1 answer:
Dahasolnce [82]3 years ago
4 0

Answer: the concentration of C6H5 NH3+ = 0.031M

Explanation:

From question pH= 5.47

pH+pOH= 14

pOH= 14-5.47= 8.53

Using the Henderson Hasselbalch equation

pOH =pKb + log (salt/base)

8.53 = 9.13 + log(salt/0.125)

8.53-9.13= log(salt/0.125)

-0.6= log(salt/0.125)

Antilog (-0.6)=salt/0.125

0.251= salt/0.125

Salt conc.= 0.125×0.251=0.031M

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Answer:

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Moles of hydregn pervade in the solution :

=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

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Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
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erastovalidia [21]

Answer:

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