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luda_lava [24]
3 years ago
8

An aqueous solution of calcium hydroxide is standardized by titration with a 0.183 M solution of hydrochloric acid. If 16.6 mL o

f base are required to neutralize 25.8 mL of the acid, what is the molarity of the calcium hydroxide solution?
Chemistry
2 answers:
Rudik [331]3 years ago
8 0

Answer:

Explanation:

Molarity of calcium hydroxide (C1) = ?

Molarity of hydrochloric acid (C2) = 0.183 M

Volume of hydrochloric acid (V2)= 25.8 ml

Volume of calcium hydroxide (V1)= 16.6 ml

Using the equation of

C1×V1 = C2×V2

C2 = 0.183 × 25.8/16.6

C2 = 0.284 M

The molarity of calcium hydroxide is 0.284 M

monitta3 years ago
5 0

Answer:

0.142 M

Explanation:

Let's consider the neutralization reaction between calcium hydroxide and hydrochloric acid.

Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O

25.8 mL of 0.183 M HCl are used. The reacting moles are:

0.0258 L × 0.183 mol/L = 4.72 × 10⁻³ mol

The molar ratio of Ca(OH)₂ to HCl is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 4.72 × 10⁻³ mol = 2.36 × 10⁻³ mol

2.36 × 10⁻³ moles of Ca(OH)₂ are contained in 16.6 mL. The molarity of Ca(OH)₂ is:

M = 2.36 × 10⁻³ mol/0.0166 L = 0.142 M

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Match the salt with the acid and base used to form it in a neutralizing reaction. K2SO4
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1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.

2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.

H₃PO₄  + 3KOH  → K₃PO₄ + 3H₂O

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At a given temperature, the elementary reaction A ---&gt;B in the forward direction is first order in A with a rate constant of
Pani-rosa [81]

Answer:

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1}   }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3}  } =5.81 \times 10^{-4}

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3 years ago
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