titration of 50ml of 0.25ammonia with 0.5M HCL .
a) find the initial PH value
b) volume of HCL is required to reach the equivalent poinnt
c)the ph at the equivalence point
4 answers
I'll help you through this but please explain what you know to do and especially what you don't understand about the problem.
i don't understand what i have to do for the calculation. what ive learnes is about acid and base titration . when we add a base to an acid until the equivalence point is reached which is where the moles of acid equals the moles of base ion .from my logicall thinking the initial ph is we use -log(H+)
and the volume of hcl is 7.
and the volume of hcl is 7.
HCl + NH3 ==> NH4Cl
b) millimols NH3 = M x mL = approx 12.5 but you need to confirm this an all of the calculations that follow.
Then millimols HCl needed = 12.5
M HCl = millimols/mL = 0.5. Solve for mL = mmols/M = 12.5/0.5 = about 25 mL.
a) You have pure NH3 at the beginning. Make an ICE chart.
...........NH3 +H2O ==> NH4^+ + OH^-
I........0.25............0.......0
C..........-x............x.......x
E......0.25-x............x........x
Substitute the E line into the Kb expression and solve for x = (OH^-) and convert that to pH.
c) You have the salt NH4Cl at the equivalence point and the pH of the solution at that point is determined by the hydrolysis of the salt, and particularly the NH4^+. (NH4Cl) at this point is millimols/mL = 12.5/(50+25) = about 0.17 and you utilize another ICE chart.
..........NH4^+ + H2O ==> H3O^+ + NH3
I.......0.17...............0.......0
C.........-x...............x.......x
E.......0.17-x.............x.......x
Substitute the E line into the Ka expression for NH4^+ (which looks like this)
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.17-x) and solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck. In the future if you post a question it helps if you tell us exactly what you don't understand about the problem. Often we can show that one step and you're on your way and we don't need to work the entire problem. That makes it easier to help more people.
b) millimols NH3 = M x mL = approx 12.5 but you need to confirm this an all of the calculations that follow.
Then millimols HCl needed = 12.5
M HCl = millimols/mL = 0.5. Solve for mL = mmols/M = 12.5/0.5 = about 25 mL.
a) You have pure NH3 at the beginning. Make an ICE chart.
...........NH3 +H2O ==> NH4^+ + OH^-
I........0.25............0.......0
C..........-x............x.......x
E......0.25-x............x........x
Substitute the E line into the Kb expression and solve for x = (OH^-) and convert that to pH.
c) You have the salt NH4Cl at the equivalence point and the pH of the solution at that point is determined by the hydrolysis of the salt, and particularly the NH4^+. (NH4Cl) at this point is millimols/mL = 12.5/(50+25) = about 0.17 and you utilize another ICE chart.
..........NH4^+ + H2O ==> H3O^+ + NH3
I.......0.17...............0.......0
C.........-x...............x.......x
E.......0.17-x.............x.......x
Substitute the E line into the Ka expression for NH4^+ (which looks like this)
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.17-x) and solve for x = (H3O^+) and convert to pH.
Post your work if you get stuck. In the future if you post a question it helps if you tell us exactly what you don't understand about the problem. Often we can show that one step and you're on your way and we don't need to work the entire problem. That makes it easier to help more people.
calculate the pH before the addition of HCL
2 answers