A 10.0-mL solution of 0.30 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution:
a) 0.00 mL
b) 10.0 mL
c) 30.0 mL
d) 40.0 mL
3 answers
Try the equation, ICE chart, etc just like the NaOH + CH3COOH problem you did previously. Then figure what you have in solution an solve accordingly. Post your work if you get stuck and I'll help you through it.
I don't even know where to start. There is no volume for HCl and am not sure how the equation is suppose to look like. I'm just confused about this problem.
Yes, there is a volume for HCl; in fact, that's what the problem is all about. That is, what is the pH after the addition of each of those quantities of 0.1M HCl.
You have 10.00 mL of 0.3M NH3 initially which is 3. mmole.
After 0 mL, that is 0 mmole HCl added.
After 10 mL x 0.1M HCl = 1 mmol added.
After 30 mL x 0.1M HCl = 3 mmole added.
After 40 mL x 0.1M HCl = 4 mmole added.
...............NH3 + HCl ==> NH4Cl
initial........3.....0.......0
added...............1............
added...............2...........
added...............3...........
added...............4...........
equil..........you can complete this.
So look to see where you are at each addition.
initially (at zero mLHCl) you have 10.0 mL of 0.3M NH3. And that's all except for the water. You know NH3 is a weak base, so you write the ionization, Kb, and solve for the OH^- and from there the pOH and pH.
At 30 mL, you have the equivalence point so all of the NH3 has been titrated with HCl and you have NH4Cl so the pH is determined by the hydrolysis of the NH4Cl salt.
All points between the beginning and the equivalence point is a buffer solution and the pH is determined by the Henderson-Hasselbalch equation.
At all point after the equivalence point, the pH is determined by the amount of excess HCl present.
Just complete the ICE chart for each of the quantities of HCl added and plug into those expressions as outlined above.
You have 10.00 mL of 0.3M NH3 initially which is 3. mmole.
After 0 mL, that is 0 mmole HCl added.
After 10 mL x 0.1M HCl = 1 mmol added.
After 30 mL x 0.1M HCl = 3 mmole added.
After 40 mL x 0.1M HCl = 4 mmole added.
...............NH3 + HCl ==> NH4Cl
initial........3.....0.......0
added...............1............
added...............2...........
added...............3...........
added...............4...........
equil..........you can complete this.
So look to see where you are at each addition.
initially (at zero mLHCl) you have 10.0 mL of 0.3M NH3. And that's all except for the water. You know NH3 is a weak base, so you write the ionization, Kb, and solve for the OH^- and from there the pOH and pH.
At 30 mL, you have the equivalence point so all of the NH3 has been titrated with HCl and you have NH4Cl so the pH is determined by the hydrolysis of the NH4Cl salt.
All points between the beginning and the equivalence point is a buffer solution and the pH is determined by the Henderson-Hasselbalch equation.
At all point after the equivalence point, the pH is determined by the amount of excess HCl present.
Just complete the ICE chart for each of the quantities of HCl added and plug into those expressions as outlined above.
1 answer