Write the chemical reaction of CH_3CH_2NH_2 reacting with HCI here. Predict how
ID: 487445 • Letter: W
Question
Write the chemical reaction of CH_3CH_2NH_2 reacting with HCI here. Predict how many ml of HCI is needed to reach the equivalence point. Explain your reasoning and show your work. At the equivalence point _ of acid added equals the original _ of base present. Predict pH after you have added a volume of X/4 of HCI (circle answer) Predict pH after you have added a volume of X/2 of HCI (circle answer). If you can figure out the answer without an ICE table, just write the answer, but explain that shortcut in words.Explanation / Answer
3) The molecular equation of the reaction is
CH3CH2NH2 + HCl ------> CH3CH2NH3+ + Cl-
We are supposed to omit the spectator ion (Cl- here) and the balanced reaction is
CH3CH2NH2 + HCl ------> CH3CH2NH3+ (ans).
4) The mL of HCl needed to reach the equivalence point will be the same as the mL of CH3CH2NH2, provided both are of the same molar concentration.
Reasoning: At the equivalence point, the number of moles of acid added equals the original number of moles of base present.
Work and prediction:
Let us assume we have V mL of x M CH3CH2NH2. Therefore, millimoles of CH3CH2NH2 present = (V mL)*(x mol/L) = Vx mmole.
As per the balanced stoichiometric reaction above,
1 mole CH3CH2NH2 = 1 mole HCl.
Therefore, Vx mmole CH3CH2NH2 = Vx mmole HCl.
Assume that we had x M HCl to start with and let V’ mL be the volume required.
Using the dilution law,
Vx = V’x
This is only possible when V’ = V (ans).
5) To answer this question, note the pKb of CH3CH2NH2 which is 3.35.
Assume we start with X mL of C M CH3CH2NH2 and we titrate the same with C M HCl.
Millimoles of CH3CH2NH2 present = XM.
We add X/4 mL of C M HCl so that millimoles of HCl added = XC/4
Set up the ICE chart as below:
CH3CH2NH2 + HCl ----> CH3CH2NH3+
initial XC XC/4 0
change -XC/4 -XC/4 + XC/4
equilibrium (XC – XC/4) 0 XC/4
The millimoles of CH3CH2NH3+ formed = millimoles of HCl added since CH3CH2NH3+ is only formed by the action of HCl on CH3CH2NH2.
Total volume of solution = (X + X/4) mL = 5X/4 mL.
[CH3CH2NH2]eq = (3XC/4)/(5X/4) = (3C/5)
[CH3CH2NH3+]eq = (XC/4)/(5X/4) = (X/5)
Use the Henderson-Hasslebach equation as
pOH = pKb + log [CH3CH2NH3+]/[CH3CH2NH2] = 3.35 + log [(X/5)/(3X/5)] = 3.35 + log (1/3)
===> p OH = 3.35 + (-0.477) = 2.873 (we find out pOH since CH3CH2NH2 is a base).
pH = 14 – pOH = 14 – 2.873 = 11.127 (ans).
6) Consider the same reaction as above. We started with XC mmole CH3CH2NH2 and we added XC/2 mmole HCl.
Mmole of CH3CH2NH3+ formed at equilibrium = XC/2 and mmoles of CH3CH2NH2 present at equilibrium = (XC – XC/2) = XC/2.
Total volume of solution = (X + X/2) mL = 3X/2 mL.
[CH3CH2NH3+]eq = (XC/2)/(3X/2) = X/3
[CH3CH2NH2]eq = (XC/2)/(3X/2) = X/3
Use the Henderson-Hasslebach equation:
pOH = pKb + log [CH3CH2NH3+]/[CH3CH2NH2] = 3.35 + log (X/3)/(X/3) = 3.35 + log (1) = 3.35.
pH = 14 – pOH = 14 – 3.35 = 10.65 (ans).
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