Answer:
Explanation:
<u>1) Data:</u>
Base: NaOH
Vb = 15.00 ml = 15.00 / 1,000 liter
Mb = ?
Acid: H₂SO₄
Va = 17.88 ml = 17.88 / 1,000 liter
Ma = 0.1053
<u>2) Chemical reaction:</u>
The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:
- Acid + Base → Salt + Water
- H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)
<u>3) Balanced chemical equation:</u>
- H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)
Placing coefficient 2 in front of NaOH and H₂O balances the equation
<u>4) Stoichiometric mole ratio:</u>
The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:
- 1 mole H₂SO₄ : 2 mole NaOH
<u>5) Calculations:</u>
a) Molarity formula: M = n / V (in liter)
⇒ n = M × V
b) Nunber of moles of acid:
- nₐ = Ma × Va = 0.1053 (17.88 / 1,000)
c) Number of moles of base, nb:
- nb = Mb × Vb = Mb × (15.00 / 1,000)
d) At equivalence point number of moles of acid = number of moles of base
- 0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)
- Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M
Answer:
10.10
Explanation:
Step 1: Write the basic dissociation reaction for pyridine
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.9 × 10⁻⁹
Step 2: Calculate [OH⁻]
For a weak base, we will use the following expression.
[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M
Step 3: Calculate pOH
We will use the definition of pOH.
pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9
Step 4: Calculate pH
We will use the following expression.
pH = 14 - pOH = 14 - 3.9 = 10.10
[A]0= Initial concentration
t1/2= half life
[A]= final concentration
k= rate constant
Answer:
The concentration of HI present at equilibrium is 0.471 M.
Explanation:
Answer:
you sure it is the right sign your using
