I'm not to sure but try soy
Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
<em>That means, 1 mole of hydroxide reacts per mole of acid</em>
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
<h3>27.6mL of LiOH 0.250M</h3>
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Answer: short answer short 0.65 / 5.37
15% NaOH by mass is 15g of NaOH in (100-15)g of solution.
No. Of moles of NaOH in solution = 0.65
No of moles of Water in solution = 4.72
Mole Fraction =
(
(Moles of Solvent)
(Moles Of Solute)+(Moles of Solvent)
)
OR
(
(Moles of Solvent)
Total Number Of Moles
)
= 0.65 / 5.37
= 0.121
Explanation:
Hope this helped
