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3 years ago

Methylamine (ch3nh2) is a weakly basic compound. calculate the kb for methylamine if a 0.253 m solution is 4.07% ionized.

1 answer:

8 0

Answer is: Kb for methylamine is 4.37·10⁻⁴.
Chemical reaction: CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻.
c(CH₃NH₂) = 0.253 M.
α = 4.07% ÷ 100% = 0.0407.
[CH₃NH₃⁺] = [OH⁻] = c(CH₃NH₂) · α.
[CH₃NH₃⁺] = [OH⁻] = 0.253 M · 0.0407.
[CH₃NH₃⁺] = [OH⁻] = 0.0103 M.
[CH₃NH₂] = 0.253 M - 0.0103 M.
[CH₃NH₂] = 0.2427 M.
Kb = [CH₃NH₃⁺] · [OH⁻] / [CH₃NH₂].
Kb = (0.0103 M)² / 0.2427 M.
Kb = 4.37·10⁻⁴.

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3 years ago

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

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Answer: The nuclear binding energy of the given element is $2.938\times 10^{12}J/mol$

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We are given:

$m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu$

M = mass of nucleus = $(n_p\times m_p)+(n_n\times m_n)$

$M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu$

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$\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol$

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So, $0.032694g/mol=0.032694\times 10^{-3}kg/mol$

To calculate the nuclear binding energy, we use Einstein equation, which is:

$E=\Delta mc^2$

where,

E = Nuclear binding energy = ? J/mol

$\Delta m$ = Mass defect = $0.032694\times 10^{-3}kg/mol$

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Putting values in above equation, we get:

$E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol$

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3 years ago

Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations:

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Answer:

(A) 1.962x10^-10 M solubility in pure water

(B) 4.0 x 10^-33 M solubility

(C) 4.0 x 10^-27 M solubility

Explanation:

(A) Fe(OH)3 would give (Fe3+) and (3OH-)

Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

Let y = [Fe^3+]

Let 3y = [OH-]

4x10^-38 = (y)(3y)^3

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y^4 = 4x10^-38 ÷ 27

y^4 = 1.481 x 10^-39

y = 1.962x10^-10 M solubility in pure water

(B) pH = 5.0

5.0 = - log [OH-]

-5.0 = log [OH-]

[OH-] = 10^-5.0 = 1.0 x 10^-5 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5

= 4.0 x 10^-33 M solubility

11.0 = - log [OH-]

-11.0 = log [OH-]

[OH-] = 10^-11.0 = 1.0 x 10^-11 M

So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38

[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38

[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11

= 4.0 x 10^-27 M solubility

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