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+

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

it can be said that

From the question we are told

A 100 mL sample of 0.25 M is titrated with a 100 mL of 0.25 M .

Therefore,

salt concentration = moles / total volume

main components : CH3NH3+

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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⁻

b.false that your answer

i believe the answer is 5,000