Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added. (The Kb for NH3 is 1.8e-05.)
(a) 0 mL
.
(b) 25.0 mL
.
(c) 59.0 mL
.
(d) 60.0 mL
(e) 63.0 mL
(f) 67.5 mL
Oh boy, let's dive into some chemistry! Don't worry, I won't make it too boring.
To calculate the pH, we need to consider the ionization of NH3 when it reacts with HCl. NH3 is a base, so it can accept a proton (H+) from HCl. The balanced equation for this reaction is:
NH3 + HCl -> NH4+ + Cl-
Now, let's break it down by volume:
(a) 0 mL: No titrant has been added yet, so the pH is just the pOH of NH3, which we can calculate using the equation:
pOH = -log(Kb) = -log(1.8e-5) = 4.74
Since pH + pOH = 14, the pH at 0 mL is:
pH = 14 - pOH = 14 - 4.74 = 9.26
(b) 25.0 mL: We need to calculate the moles of HCl that have reacted with NH3. Since they have a 1:1 mole ratio, the moles of HCl is equal to the moles of NH3 that reacted. To find the moles:
moles NH3 = (volume of NH3) x (concentration of NH3) = (30.0 mL) x (0.050 M) = 1.5 mmol
The moles of HCl is also 1.5 mmol because of the 1:1 mole ratio. Now, we need to calculate the concentration of NH3 and HCl after the reaction:
moles NH3 remaining = moles NH3 initial - moles NH3 reacted = 1.5 mmol - 1.5 mmol = 0 mmol
The volume after adding 25.0 mL is:
volume = (initial volume) + (volume added) = 30.0 mL + 25.0 mL = 55.0 mL
Now, we can calculate the concentration of NH4+ (formed from NH3) using the new volume:
concentration NH4+ = (moles NH4+) / (volume) = (1.5 mmol) / (55.0 mL) = 0.027 M
Since NH4+ is a weak acid, we need to calculate the pH using the equation:
pH = pKa + log([concentration NH4+] / [concentration NH3])
For NH4+, pKa = -log(Kb) = -log(1.8e-5) = 4.74
Now, let's substitute the values:
pH = 4.74 + log(0.027 M / 0.050 M) = 4.74 + log(0.54) = 4.74 + (-0.27) = 4.47
(c) 59.0 mL: We'll follow the same steps as in (b), but with a new volume of 59.0 mL.
volume = 30.0 mL + 59.0 mL = 89.0 mL
concentration NH4+ = (moles NH4+) / (volume) = (1.5 mmol) / (89.0 mL) = 0.017 M
pH = 4.74 + log(0.017 M / 0.050 M) = 4.74 + log(0.34) = 4.74 + (-0.47) = 4.27
(d) 60.0 mL: Same as (c), but with a volume of 60.0 mL.
volume = 30.0 mL + 60.0 mL = 90.0 mL
concentration NH4+ = (moles NH4+) / (volume) = (1.5 mmol) / (90.0 mL) = 0.0167 M
pH = 4.74 + log(0.0167 M / 0.050 M) = 4.74 + log(0.334) = 4.74 + (-0.474) = 4.27
(e) 63.0 mL: Same as (c), but with a volume of 63.0 mL.
volume = 30.0 mL + 63.0 mL = 93.0 mL
concentration NH4+ = (moles NH4+) / (volume) = (1.5 mmol) / (93.0 mL) = 0.0161 M
pH = 4.74 + log(0.0161 M / 0.050 M) = 4.74 + log(0.322) = 4.74 + (-0.491) = 4.25
(f) 67.5 mL: Same as (c), but with a volume of 67.5 mL.
volume = 30.0 mL + 67.5 mL = 97.5 mL
concentration NH4+ = (moles NH4+) / (volume) = (1.5 mmol) / (97.5 mL) = 0.0154 M
pH = 4.74 + log(0.0154 M / 0.050 M) = 4.74 + log(0.308) = 4.74 + (-0.512) = 4.23
Phew! I hope that was helpful, and not too acidic for your taste. If you have any more questions, feel free to ask!
To calculate the pH at different points during the titration, we need to determine the moles of NH3 and HCl at each volume of titrant added. Then, we can use the Henderson-Hasselbalch equation to calculate the pH. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (in this case, pKa = -log(Kb)), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
Given information:
Initial volume of NH3 (acid) = 30.0 mL
Concentration of NH3 (acid) = 0.050 M
Concentration of HCl (titrant) = 0.025 M
Kb for NH3 (base) = 1.8e-05
Let's calculate the pH at each point:
(a) 0 mL titrant added:
At this point, we have only NH3 (acid) in the solution, so it is a pure acid solution. The moles of NH3 = concentration x volume = 0.050 M x 30.0 mL = 1.5 mmol. Since there is no titrant added, the moles of HCl = 0 mmol. Therefore, [A-] = 0 and [HA] = 1.5 mmol. Since Kb = [A-][H+]/[HA], and [A-] = 0, the concentration of [H+] is also 0. Taking the negative logarithm of 0 gives us a pH of infinity.
(b) 25.0 mL titrant added:
At this point, we have added 25.0 mL of HCl to the solution. The moles of NH3 = 0.050 M x 30.0 mL = 1.5 mmol (unchanged). The moles of HCl = 0.025 M x 25.0 mL = 0.625 mmol. The excess HCl will react with the NH3 to form NH4+ ions, leaving some NH3 unreacted. The remaining moles of NH3 = 1.5 mmol - 0.625 mmol = 0.875 mmol. Therefore, [A-] = 0.625 mmol and [HA] = 0.875 mmol. Now we can calculate the [H+]:
Kb = [A-][H+]/[HA]
1.8e-05 = (0.625 mmol)([H+])/(0.875 mmol)
[H+] = (1.8e-05)(0.875 mmol)/0.625 mmol ≈ 2.52e-05 M
Taking the negative logarithm of [H+] gives us the pH:
pH = -log(2.52e-05) ≈ 4.60
(c) 59.0 mL titrant added:
At this point, we have added 59.0 mL of HCl. The moles of NH3 and HCl remain the same, but now we have added enough HCl to react with all the NH3. Therefore, all the NH3 has been converted to NH4+ ions, and [A-] = 1.5 mmol and [HA] = 0 mmol. Now we can calculate the [H+]:
Kb = [A-][H+]/[HA]
1.8e-05 = (1.5 mmol)([H+])/0
Since [HA] = 0, [H+] is undefined and the solution is neutral. The pH is 7.
(d) 60.0 mL titrant added:
At this point, we have added 60.0 mL of HCl. The moles of NH3 and HCl remain the same, but HCl is in excess now. Therefore, all the NH3 has been converted to NH4+ ions, and [A-] = 1.5 mmol and [HA] = 0 mmol. Similar to (c), the [H+] is undefined, and the solution is neutral. The pH is 7.
(e) 63.0 mL titrant added:
At this point, we have added 63.0 mL of HCl. The moles of NH3 and HCl remain the same, but HCl is still in excess. Therefore, all the NH3 has been converted to NH4+ ions, and [A-] = 1.5 mmol and [HA] = 0 mmol. Similar to (c) and (d), the [H+] is undefined, and the solution is neutral. The pH is 7.
(f) 67.5 mL titrant added:
At this point, we have added 67.5 mL of HCl. The moles of NH3 remain the same, but HCl is still in excess. Therefore, all the NH3 has been converted to NH4+ ions, and [A-] = 1.5 mmol and [HA] = 0 mmol. Similar to (c), (d), and (e), the [H+] is undefined, and the solution is neutral. The pH is 7.
Summary of pH values at different volumes of titrant added:
(a) 0 mL - pH is undefined
(b) 25.0 mL - pH ≈ 4.60
(c) 59.0 mL - pH is 7
(d) 60.0 mL - pH is 7
(e) 63.0 mL - pH is 7
(f) 67.5 mL - pH is 7
To calculate the pH after different volumes of titrant have been added in the titration of NH3 with HCl, we need to consider the reaction that takes place between NH3 and HCl. The balanced chemical equation for this reaction is:
NH3 + HCl -> NH4+ + Cl-
The NH3 acts as a base and the HCl acts as an acid. NH3 accepts a proton from HCl to form NH4+ (ammonium ion), while Cl- is the conjugate base of HCl.
To determine the pH at different volumes of titrant added, we need to calculate the concentration of NH4+ and then determine the concentration of OH- (which is produced by the hydrolysis of NH4+) using the Kb value for NH3. Finally, we can use the concentration of OH- to calculate the pH.
Let's go through each volume of titrant step by step:
(a) 0 mL:
No titrant has been added yet, so the NH3 is in its original concentration. The concentration of NH3 is 0.050 M. Since no OH- has been produced, the pH will be determined by the hydrolysis of the NH4+ ion. The Kb for NH3 is 1.8e-5. We can use this value to calculate the concentration of OH- and then use that to determine the pH.
(b) 25.0 mL:
At 25.0 mL, half the volume of titrant has been added. We can calculate the moles of NH4+ formed by reacting 25.0 mL of 0.025 M HCl with NH3. From the balanced equation, we know that one molecule of NH3 reacts with one molecule of HCl to form one molecule of NH4+. Since NH3 is in excess, the moles of NH4+ formed will be equal to the moles of HCl reacted. Determine the moles of NH4+ formed and then calculate its concentration by dividing by the total volume of the resulting solution. Use the Kb value to calculate the concentration of OH- and then the pH.
(c) 59.0 mL, (d) 60.0 mL, (e) 63.0 mL, and (f) 67.5 mL:
Repeat the same process as in step (b), but now consider the moles of NH4+ formed by reacting the given volume of titrant with NH3. Calculate the concentration of NH4+ and then determine the concentration of OH- using the Kb value. Finally, calculate the pH using the concentration of OH-.
Remember to keep track of the total volume of the resulting solution as more titrant is added, as this will affect the concentration calculations.
By following these steps, you can calculate the pH at each given volume of titrant added in the titration of NH3 with HCl.
NH3 + HCl ==> NH4Cl
First, calculate the volume HCl needed for the equivalence point. Then divide the "curve" into these portions. You need to where you are on the curve to know how to calculate the pH.
a. beginning point. This is just a weak base, NH3. You know Kb and (NH3), use the
NH3 + HOH ==> NH4^+ + OH^-
Set up an ICE chart and solve for x = OH^- and convert to pH.
b. ALL point between the zero mL and the equivalence point. Use the Henderson-Hasselbalch equation.
c. The equivalence point, Set up an ICE chart for the salt produced at the equivalence, hydrolyze the NH4^+ and use Ka NH4^+ (rememberr that's Kw/Kb for NH3)
d. For all points after the equivalence point use the concn of the excess HCl and convert to pH.