1. a Lewis acid in water- it accepts a share of an electron pair from a water mo

Question

1. a Lewis acid in water- it accepts a share of an electron pair from a water molecule, and CO2 gas acts as forms carbonic acid, H,Co, Carbonic acid is a weak acid: K, 42x 107. If you open a cold, carbonated beverage like sparkling water, what will happen to releases CO, gas? (No calculation required. Use chemical equilibria equations to support your answer) the pH of the beverage as it warms and 2. The pK, of HF (hydrofluoric acid) is 3.5. Calculate the pH of 250 mL of a solution that contains 0.2 moles of HF and 0.3 moles of NaF using the H-H equation. 3. Calculate the ratio of HF to NaF that would be required to yield a solution with a pH of 3.24. The pK, for HF is 3.5 4. Soon in lab you'll explore the titration of the weak base NH3 (ammonia) with the strong acid HCI The Kp for NH3 is 1.8 x 105, Calculate the pH of a 0.1 M aqueous solution of NH i.e. before any HCl is added in your titration) If you titrated 50 mL of 0.1 M NH, to the point where 25 mL of 0.1 M HCI had been added, what would the pH of the solution be? If you titrated 50 mL of 0.1 M NH, to the point where 50 mL of 0.1 M HCI had been added, what would th pH of the solution be?

Explanation / Answer

2.

Molarity of HF = 0.2 mol / 0.25 L = 0.8 M

Molarity of NaF = 0.3 mol / 0.25 L = 1.2 M

Using Henderson-hessalbalach equation

pH = pKa + log { [salt] / [acid] }
      = pKa + log { [NaF] / [HF] }
      = 3.5 + log { 1.2 / 0.8 }
    = 3.5 + log (1.5)
    = 3.5 + 0.18
    = 3.68

3.

Using Henderson-hessalbalach equation

pH = pKa + log { [salt] / [acid] }
pH = pKa + log { [NaF] / [HF] }
3.24 = 3.5 + log { [NaF] / [HF] }
log { [NaF] / [HF] } = - 0.26
[NaF] / [HF] = 10-0.26
[NaF] / [HF] = 0.55
[HF] / [NaF] = 1 /0.55
[HF] / [NaF] = 1.82/1

4.

                        NH3           +            H2O               <-------->          NH4+            +               OH-
IC:                   0.1                                                                              0                                    0
C:                    - x                                                                              +x                                 +x
EC:               0.1 – x                                                                           x                                    x

Kb = [NH4+] [OH-] / [NH3]
or, 1.8 x 10-5 = (x) (x) / (0.1 – x)
or, 1.8 x 10-5 = x2 / (0.1 – x)
or, 1.8 x 10-5 = x2 / (0.1)
or, x2 = 1.8 x 10-6
or, x = 1.34 x 10-3

So, [OH-] = x = 1.34 x 10-3 M

pOH = -log [OH-]
        = - log (1.34 x 10-3)
        = 2.87

pH = 14 – pOH
     = 14 – 2.87
     = 11.13

(b)

50 mL of 0.1 M NH3
Moles of NH3 = 0.1 M x 0.050 L = 0.005 moles

25 mL of 0.1 M HCl
Moles of NH3 = 0.1 M x 0.025 L = 0.0025 moles

0.0025 moles of HCl will react with 0.0025 moles of NH3 to produce 0.0025 moles of NH4+. So, moles of NH3 left = 0.0025 moles

Total volume = 50 mL + 25 mL = 75 mL = 0.075 L

Now, the concentration terms are
[NH4+] = 0.0025 mol / 0.075 L = 0.033 M
[NH3] = 0.0025 mol / 0.075 L = 0.033 M

Using Henderson-hessalbalach equation

pOH = pKb + log { [salt] / [acid] }
pOH = pKb + log { [NH4+] / [NH3] }
pOH = pKb + log { 0.033 / 0.033 }
pOH = pKb + log { 1 }
pOH = pKb + 0
pOH = pKb

Kb of NH3 = 1.8 x 10-5

pKb = - log (1.8 x 10-5) = 4.74

So, pOH = pKb = 4.74

pH = 14 – pOH = 14 – 4.74 = 9.26

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.