Question

A 100 mL sample of 0.25 M CH3NH2(aq) is titrated with a 100 mL of 0.25 M HNO3(aq). Select ALL main components (greater than 0.001 moles, besides H2O) that would be present in the solution after adding HNO3. Kb of CH3NH2 is 4.4 LaTeX: \times×10−4.
OH−
CH3NH2
NO3−
CH3NH3+

Answer 1

Answer:

The components present in the solution are:

CH₃NH₂

CH₃NH₃⁺

NO₃⁻

Explanation:

CH₃NH₂(aq) + HNO₃(aq) → CH₃NH₃⁺ NO₃⁻

t = 0,

milliequivalent

100*0.25 = 25   50*0.25 = 12.5

End of reaction:

25 - 12.5 = 12.5,  12.5 - 12.5 = 0,  25 - 12.5 = 12.5

moles of CH₃NH₂ left    = 12.5x10⁻³ = 0.0125 > 0.001

moles of HNO₃ left        = 0

moles of CH₃NH₃⁺ left  = 12.5x10⁻³ = 0.0125 > 0.001

moles of NO₃⁻ left         = 12.5x10⁻³ = 0.0125 > 0.001

Therefore, the component present in the solution are: CH₃NH₂ , CH₃NH₃⁺ , NO₃⁻

Answer 2

We have that for the Question "Select ALL main components (greater than 0.001 moles, besides$H_2O$) that would be present in the solution after adding $HNO_3$."

it can be said that

• $CH_3NH_3+$ would be present in the solution after adding $HNO_3$

From the question we are told

A 100 mL sample of 0.25 M $CH_3NH_2(aq)$ is titrated with a 100 mL of 0.25 M $HNO_3(aq)$.

Therefore,

$moles of = 100 * 0.25 = 25\\\\moles of HNO_3 = 100 * 0.25 = 25$

$Kb = 4.4 x 10^{-4}\\\\pKb = 3.36$

$CH_3NH_2 (aq) + HNO_3(aq) ---------> CH_3NH_3+$

salt concentration = moles / total volume

$= \frac{25}{100 + 100}\\\\= 0.125 M$

$pH = 7 - \frac{1}{2} (pKb + log C)\\\\= 7 - \frac{1}{2} (3.36 + log 0.125)\\\\= 5.77$

main components : CH3NH3+

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