Answer:
(a) 2.7
(b) 4.44
(c) 4.886
(d) 5.363
(e) 5.570
(f) 12.30
Explanation:
Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.
Lets call HA propanoic acid and A⁻ its conjugate base,
(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7
(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol
mol left HA = 0.28 - 0.007 = 0.021
mol A⁻ produced = 0.007
Using the Hasselbalch-Henderson equation for buffer solutions:
pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44
(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol
mol HA left = 0.028 -0.014 = 0.014 mol
mol A⁻ produced = 0.014
pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886
(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol
mol HA left = 0.028 - 0.021 = 0.007
mol A⁻ produced = 0.021
pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363
(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol
mol HA left = 0
Now we only a weak base present and its pH is given by:
pH = √(kb x (A⁻) where Kb= Kw/Ka
Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)
mol A⁻ = 0.028 mOl
Vol solution = 100 mL + 200 mL = 300 mL
(A⁻) = 0.028 mol /0.3 L = 0.0093 M
and we also need to calculate the Kb for the weak base:
Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰
pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570
(f) Treat this part as a calculation of the pH of a strong base
moles of OH = 0.250 L x 0.14 mol = 0.0350 mol
mol OH remaining = 0.035 mol - 0.028 reacted with HA
= 0.007 mol
(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²
pOH = - log (2.00 x 10⁻²) = 1.70
pH = 14 - 1.70 = 12.30
)
= 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
