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3241004551 [841]
3 years ago
9

An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t

he base solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added. g
Chemistry
1 answer:
Elenna [48]3 years ago
4 0

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol

Thus, the concentration of ethylamine in solution is:

[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M

Now, we can also infer that some salt is formed, and has the following concentration:

[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0

Finally, the pH turns out to be:

pH=14-pOH=14-3\\\\pH=11

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

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Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

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<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

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Also, Given that:

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Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

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So,  

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Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

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moles = \frac{Mass\ taken}{Molar\ mass}

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Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

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So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

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