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almond37 [142]
4 years ago
7

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Chemistry
1 answer:
Andru [333]4 years ago
3 0

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

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Explanation:

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Question 5 (1 point)
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Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

 n= number of moles

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 T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

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3 years ago
Calculate the thermometer in an air-conditioned room reads 20.0°C what is the temperature of the room In degrees Fahrenheit and
maks197457 [2]

Answer:

A. 68°F

B. 293K

Explanation:

The temperature (celsius) = 20.0°C

A. Conversion of the temperature in celsius to fahrenheit.

This is illustrated below:

°F = 9C/5 + 32

C = 20°C

°F = 9C/5 + 32

°F = 9x20/5 + 32

°F = 36 + 32

°F = 68°F

B. Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 20°C

Temperature (Kelvin) = 20°C + 273

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5 0
3 years ago
An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t
Elenna [48]

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!

4 0
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Balanced equation including phases for the reaction of hydroiodic acid and lithium hydrosulfite
leva [86]
And so what are you trying to ask or are you making a statement??????????????!☺☺☺
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