What is the pH at the point during a titration when 34 mL of 0.200 M HCl has been added to 29 mL of 0.0500 M NH3 solution? HCl is a strong acid, Kb=1.8×10−5 for NH3.
Answer in units of pH
millimols HCl = 34 x 0.2 = 6.8
mmols NH3 = 29 x 0.05 = 1.45
..........HCl + NH3 ==> NH4Cl
I.........6.8...1.45......0
C........-1.45.-1.45.....+1.45
E.........5.35...0........1.45
If you look at the numbers you will notice that the titration with HCl has neutralized all of the NH3 and the point in the titration is well above the equivalence point. So you have excess HCl in NH4Cl.
(H^+) = excess HCl and pH from that.
(HCl) = mmols/mL = 5.35/63 = ?
To determine the pH at a specific point during a titration, we need to consider the reaction between the acid and the base and calculate the concentration of the resulting species.
In this case, HCl is a strong acid and completely dissociates in water, while NH3 is a weak base with a specified Kb value. The reaction between HCl and NH3 can be represented as follows:
HCl + NH3 -> NH4+ + Cl-
To find the pH, we need to determine the concentration of the NH4+ ion. This can be done using the concept of stoichiometry and the initial concentrations of the solutions.
Given the initial volume and concentration of the HCl solution (34 mL and 0.200 M, respectively), we can calculate the number of moles of HCl initially present by multiplying the volume by the concentration:
moles of HCl = (34 mL) * (0.200 M) = 6.8 mmol
Similarly, we can calculate the number of moles of NH3 using the initial volume and concentration (29 mL and 0.0500 M, respectively):
moles of NH3 = (29 mL) * (0.0500 M) = 1.45 mmol
Since HCl and NH3 react in a 1:1 ratio, the limiting reactant is NH3. This means that all 1.45 mmol of NH3 will react to form NH4+.
Therefore, the final concentration of NH4+ is 1.45 mmol divided by the total volume of the solution (34 mL + 29 mL = 63 mL):
[ NH4+ ] = (1.45 mmol) / (63 mL) = 0.0230 M
To calculate the pOH, we can use the Kb value given for NH3:
Kb = [ NH4+ ][ OH- ] / [ NH3 ]
Since the concentration of OH- is equal to the concentration of NH4+ (as the reaction ratio is 1:1), we can substitute [ OH- ] with [ NH4+ ]:
Kb = [ NH4+ ]^2 / [ NH3 ]
Rearranging the equation, we can solve for [ OH- ]:
[ NH4+ ]^2 = Kb * [ NH3 ]
[ NH4+ ]^2 = (1.8×10^-5) * (0.0230 M)
[ NH4+ ]^2 = 4.14 × 10^-7
[ NH4+ ] = sqrt(4.14 × 10^-7) = 6.43 × 10^-4 M
From here, we can calculate pOH as follows:
pOH = -log10( [ OH- ] )
pOH = -log10( 6.43 × 10^-4 )
pOH = 3.19
To find the pH, we can use the equation:
pH = 14 - pOH
pH = 14 - 3.19
pH = 10.81
Therefore, the pH at the specified point during the titration is 10.81.