Question

In u L. methy of the What is the pH of 50 mL of a lo molle hylamine ne (CH₃NH₂) colution after the addition of some of mol/L

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Answer #1

Concentration of 50 ml , 1.0 M ethylamine (base) = 50 × 1

= 50 mM

Concentration of 50 ml , 0.5 M HCl (acid) = 50 × 0.5

= 25 mM

Here concentration of acid neutralizes concentration of base. Hence concentration of base remaining = 50 (mM) - 25 (mM)

= 25 mM

Now pOH = - log(base)

= - log ( 25×10-3)

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