Calculate the pH after 0.014 mole of gaseous HCl is added to 227.0 mL of each of the following buffered solutions. (Assume that all solutions are at 25°C.)
(a) 0.089 M NH3 and 0.17 M NH4Cl
(b) 0.74 M NH3 and 1.25 M NH4Cl
I like to work in millimols.
millimoles NH3 = mL x M = 227.0 mL x 0.089 = approx 20 but you need a better number on this and all other estimates that follow.
millimoes NH4Cl = 227.0 x 0.17 = about 39
millimols HCl added = 14
...................NH3 + HCl ==> NH4Cl
I..................20........0...............39
add.......................14.................
C............-...14......-14...............+14
E...................6........0.................53..
The problem doesn't give you the volume so use millimoles as the same as M. They aren't, of course, however, in the calculation the total volume cancels so it isn't necessary. Plut millimoles into the Henderson-Hasselbalch equation and solve for pH.
The second part is done the same way.
B+ H2O <----->BH +OH^-
Kb=[BH][OH^-]/[B]
Kb=([BH]/[B])*[OH^-]
-logKb=-log([BH]/[B])*-log[OH^-]
pKb=-log([BH]/[B])+pOH
Variation of the Henderson-Hasselbach Equation:
pOH=pKb+log[BH]/[B]
14-pOH=pH
Convert to moles for each:
0.089 M NH3*0.227L= 0.020203 moles of NH3
0.17 M*0.227L= 0.03859 moles of NH4Cl
Addition of a strong acid to a weak base in its conjugate acid will increase the amount of conjugate acid and decrease the amount of weak base by the same amounts. You can use an ICE chart if you like, which can be beneficial in visualizing what is happening.
B + HCl ---> BH +Cl
BH=NH3Cl=0.03859+0.014=0.05259
B=NH3=0.020203-0.014=0.006203
Plug into equation:
let pKb=4.75 but use the one from your text.
pOH=4.75+log[0.05259]/[0.006203]
pOH=4.75+0.9283=5.678
pH=14-5.678=8.32
B.) Is the same way.
To calculate the pH of a buffered solution, we need to determine the concentration of the components in the solution and use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
where:
pH is the logarithmic scale of the hydrogen ion concentration,
pKa is the acid dissociation constant of the weak acid (HA) in the buffer,
[A-] is the concentration of the conjugate base,
[HA] is the concentration of the weak acid.
For both parts (a) and (b) of the question, we have the same weak acid (NH4Cl) and conjugate base (NH3), but different concentrations.
(a) 0.089 M NH3 and 0.17 M NH4Cl
The pKa value for the NH4Cl/NH3 system is 9.25.
To use the Henderson-Hasselbalch equation, we need to calculate the concentrations of NH3 [A-] and NH4Cl [HA].
For NH3:
moles of NH3 = concentration (M) x volume (L)
moles of NH3 = 0.089 M x 0.227 L = 0.020203 moles
For NH4Cl:
moles of NH4Cl = concentration (M) x volume (L)
moles of NH4Cl = 0.17 M x 0.227 L = 0.03859 moles
To find the concentration of NH3 and NH4Cl, we divide the moles by the total volume in liters:
[A-] = moles of NH3 / total volume (L)
[A-] = 0.020203 moles / 0.227 L = 0.0889 M
[HA] = moles of NH4Cl / total volume (L)
[HA] = 0.03859 moles / 0.227 L = 0.170 M
Now, we can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = 9.25 + log (0.0889 / 0.170)
pH = 9.25 + log (0.5235)
pH = 9.25 + (-0.281)
pH = 8.969
Therefore, the pH of the solution is approximately 8.969.
(b) 0.74 M NH3 and 1.25 M NH4Cl
The calculations for part (b) are similar to part (a).
For NH3:
moles of NH3 = 0.74 M x 0.227 L = 0.16838 moles
For NH4Cl:
moles of NH4Cl = 1.25 M x 0.227 L = 0.28375 moles
[A-] = 0.16838 moles / 0.227 L = 0.741 M
[HA] = 0.28375 moles / 0.227 L = 1.25 M
pH = 9.25 + log (0.741 / 1.25)
pH = 9.25 + log (0.5928)
pH = 9.25 + (-0.226)
pH = 9.024
Therefore, the pH of the solution is approximately 9.024.
Note: In both cases, we assumed that the acid dissociation constant (pKa value) of NH4Cl/NH3 system is 9.25 at 25°C.