Question

Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b. (7pt) What is the pH at 6.00 mL base added? c. (8pt) What is the pH when 40.00 mL of KOH is added to reach the equivalence point? d. (8pt) What is the pH if 42.00 mL of KOH is titrated with the solution?

Answer 1

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