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8_murik_8 [283]
3 years ago
12

Butyric acid is responsible for the odor in rancid butter. A solution of 0.25 M butyric acid has a pH of 2.71. What is the Ka fo

r the acid? Multiple Choice 0.36 2.4 × 10–2 7.8 × 10–3 1.5 × 10–5
Chemistry
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

Ka = 1.5 × 10⁻⁵

Explanation:

Butyric acid is a weak acid that ionizes according to the following equation:

CH₃-CH₂-CH₂-COOH(aq) ⇄ CH₃-CH₂-CH₂-COO⁻(aq) + H⁺(aq)

We can find the value of the acid dissociation constant (Ka) using the following expression:

Ka=\frac{[H^{+}]^{2} }{Ca}

where

[H⁺] is the molar concentration of H⁺

Ca is the initial molar concentration of the acid

We can find [H⁺] from the pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.71 = 1.95 × 10⁻³ M

Then,

Ka=\frac{(1.95 \times 10^{-3})^{2} }{0.25} =1.5 \times 10^{-5}

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2 years ago
How many moles of KMnO, are in 66.38 g of KMnO?
schepotkina [342]

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5 0
3 years ago
In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters
IrinaK [193]

<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol

For the given chemical reaction:

Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = \frac{1}[1}\times 0.0851=0.0851mol of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Putting values in above equation, we get:

733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L

Hence, the volume of hydrogen gas collected over water is 2.13 L

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3 years ago
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A. Nonmetallic
B. Nonmetallic
C. Metallic
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3 years ago
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