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⁻