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
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.
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.
calculate the pH before the addition of HCL
To find the answers to the questions related to the titration of ammonia with HCl, we need to follow a step-by-step approach. Here's how to get the solution for each question:
a) Finding the initial pH value:
1. Ammonia (NH3) is a weak base, while HCl is a strong acid.
2. However, in this titration, we are using a limited volume of ammonia (50 mL) and a higher concentration of HCl (0.5 M).
3. Before any reaction occurs, the solution consists of only ammonia.
4. To find the initial pH, we can calculate the pOH of the ammonia solution.
- pOH = -log10[OH-]
- OH- concentration can be obtained by multiplying the initial concentration of ammonia (0.25 M) by the dilution factor (50 mL / 1000 mL).
- Once we have pOH, we can subtract it from 14 to get the initial pH value.
b) Finding the volume of HCl required to reach the equivalence point:
1. The equivalence point is the point at which the moles of acid and base are stoichiometrically equal.
2. For this titration, the balanced equation is NH3 + HCl -> NH4Cl.
3. To reach the equivalence point, we need to add an equal number of moles of HCl as there are moles of NH3.
4. Calculate the moles of NH3 by multiplying the initial concentration (0.25 M) by the volume (50 mL / 1000 mL).
5. Based on the balanced equation, the stoichiometric ratio between NH3 and HCl is 1:1.
6. The moles of HCl required to reach the equivalence point will be the same as the moles of NH3.
c) Finding the pH at the equivalence point:
1. At the equivalence point, all the NH3 reacts with HCl, forming NH4Cl.
2. NH4Cl is a salt formed by a strong acid (HCl) and a weak base (NH3).
3. NH4+ is the conjugate acid of NH3, and Cl- is the conjugate base of HCl.
4. The pH at the equivalence point depends on the hydrolysis of NH4Cl. We need to determine whether it is acidic, basic, or neutral.
5. For NH4Cl, NH4+ acts as an acid, and Cl- acts as a base through hydrolysis. The degree of hydrolysis depends on the equilibrium constant, Kb.
6. Kb can be calculated using the Kb expression and the concentration of NH4Cl.
7. Once we have Kb, we can calculate the pH using the formula:
- pOH = -log10[sqrt(Kb * [NH4Cl])]
- pH = 14 - pOH
Keep in mind that these calculations involve assumptions and simplifications. To obtain more precise results, accurate acid-base equilibrium constants and any other relevant factors should be considered.