the answer to the first one is D and the second one is the first one
Answer:
4.27
Explanation:
Let's consider the dissociation of a generic monoprotic acid.
HA(aq) → H⁺(aq) + A⁻(aq)
The pH is 2.36. The concentration of H⁺ is:
pH = -log [H⁺]
[H⁺] = antilog -pH
[H⁺] = antilog -2.36 = 4.37 × 10⁻³ M
We know that the concentration of the acid Ca = 0.3535 M. We can find the acid dissociation constant using the following expression.
[H⁺] = √(Ca × Ka)
Ka = [H⁺]²/Ca
Ka = (4.37 × 10⁻³)²/0.3535
Ka = 5.40 × 10⁻⁵
The pKa is:
pKa = -log Ka = -log 5.40 × 10⁻⁵ = 4.27
<span>336*280 i believe... i hope this helps
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Answer:
1.195 M.
Explanation:
- We can calculate the concentration of the stock solution using the relation:
<em>M = (10Pd)/(molar mass).</em>
Where, M is the molarity of H₂SO₄.
P is the percent of H₂SO₄ (P = 40%).
d is the density of H₂SO₄ (d = 1.17 g/mL).
molar mass of H₂SO₄ = 98 g/mol.
∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.
- We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
<em>∴ (MV) before dilution = (MV) after dilution</em>
M before dilution = 4.78 M, V before dilution = 250 mL.
M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.
∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.
