Answer:
a. pH = 4.75
b. pH = 9.20
c. pH = 8.42
d. pH = 13.53
Explanation:
This is a titration between a strong base, the KOH and a weak acid, HCN.
The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN
HCN + H2O ⇄ H₃O⁺ + CN⁻
Initial 0.5 - -
Eq. 0.5-x x x
Ka = x² / (0.5-x) = 6.17ₓ10⁻¹⁰
Ka is really small, so we can say that 0.5-x = 0.5. Then,
x² = 6.17ₓ10⁻¹⁰ . 0.5
x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]
pH = - log [H₃O⁺] → - log 1.75×10⁻⁵ = 4.75
b. First of all, we determine the moles of base, we are adding.
0.250 mol/L . 0.006 L = 0.0015 moles
In conclussion we have 0.0015 moles of OH⁻
Now, we determine the moles of our acid.
0.500 mol/L . 0.020L = 0.01 moles
The 0.0015 moles of OH⁻ will be neutralized with the acid, so:
HCN + OH⁻ → H₂O + CN⁻
0.01 0.0015 0.0085
The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)
Our new volume is 20 mL and 6mL that we added, so, 26mL
This is a buffer with the weak acid, and its conjugate base.
Our concentrations are 0.0085 moles / 0.026 L = 0.327 M
We apply Henderson-Hasselbach
pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)
pH = pKa
c. When we add 40 mL, our volume is 20mL +40mL = 60 mL
These are the moles, we add:
0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)
HCN + OH⁻ → H₂O + CN⁻
0.01 0.01 0.01
All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.
0.01 moles / 0.060 L = 0.16 M → [CN⁻]
pH at this point will be
CN⁻ + H₂O ⇄ HCN + OH⁻ Kb = 1.62ₓ10⁻⁵ (Kw/Ka)
In. 0.16 - -
Eq. 0.16-x x x
Kb = x² / (0.16-x)
We can also assume that 0.16-x = 0.16. Then:
[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ . 0.16) = 2.59×10⁻⁶
- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58
pH = 14 - pOH → 14 - 5.58 = 8.42
This is a basic pH, because the titration is between a weak acid and a strong base.
d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL
We add 0.5 mol/L . 0.062L = 0.031 moles
These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.
0.031 moles - 0.01 moles = 0.021 moles of OH⁻
[OH⁻] = 0.021 moles / 0.062L = 0.34M
- log [OH⁻] = pOH → - log 0.34 = 0.47
pH = 14-pH → 14 - 0.47 = 13.53
