Asked by Melissa
A 119.2mL sample of 0.105M methylamine (CH3NH2, Kb=3.7*10^-4) is titrated with 0.255M HNO3. Calculate the after the addition of each of the following volumes of acid. 49.1mL and 73.6mL. Please help.
Answers
Answered by
DrBob222
a). First, you omitted what you want t calculate!!!
.......CH3NH2 + HNO3 ==> CH3NH3^+ + NO3^-
mmoles CH3NH2 = 119.2 mL x 0.105 = ??
mmoles HNO3 = 49.1 x 0.255 = ??
Take a look at the numbers; I think the rounded numbers (to 3 s.f.) are equal and the pH (H^+) at the equivalence point is determined by the hydrolysis of the salt.
CH3NH3^+ + H2O ==> CH3NH2 + H3O^+
Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+)
(H3O^+) = (CH3NH2) = x
(CH3NH3^+) = moles HNO3/total mL OR moles CH3NH2/total mL.
b. 73.6 mL HNO3 is FAR past the equivalencae point; therefore, the pH at that point is determined by the excess HNO3. Use mmoles HNO3 - mmoles CH3NH2 = mmoles HNO3. Divide that by mL for final (HNO3) and calculate pH or whatever you want.
.......CH3NH2 + HNO3 ==> CH3NH3^+ + NO3^-
mmoles CH3NH2 = 119.2 mL x 0.105 = ??
mmoles HNO3 = 49.1 x 0.255 = ??
Take a look at the numbers; I think the rounded numbers (to 3 s.f.) are equal and the pH (H^+) at the equivalence point is determined by the hydrolysis of the salt.
CH3NH3^+ + H2O ==> CH3NH2 + H3O^+
Ka = (Kw/Kb) = (H3O^+)(CH3NH2)/(CH3NH3^+)
(H3O^+) = (CH3NH2) = x
(CH3NH3^+) = moles HNO3/total mL OR moles CH3NH2/total mL.
b. 73.6 mL HNO3 is FAR past the equivalencae point; therefore, the pH at that point is determined by the excess HNO3. Use mmoles HNO3 - mmoles CH3NH2 = mmoles HNO3. Divide that by mL for final (HNO3) and calculate pH or whatever you want.
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