Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:
Dimensional Analysis:
In conclusion, about 547 grams of copper (II) bicarbonate is produced.
Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0