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2 years ago

Calcula el pH de las siguientes sustancias formadas durante una erupción volcánica:

Chemistry

1 answer:

Virty [35]2 years ago

8 0

<em>Calculate the pH of the following substances formed during a volcanic eruption: </em>

<em>• Sulfurous acid if [H +] = 0.10 </em>

<em>• Nitric acid if [H +] = 0.11</em>

<em />

<h3>Further explanation </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

pH acid rain

$\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72$

pH Sulfurous acid

$\tt pH=-log[10^{-1}]\\\\pH=1$

pH Nitric acid

$\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959$

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Table 2 Heat of Reaction of NaOH (aq) and HCl(aq) Solution

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2 years ago

Potassium hydroxide is classified as an arrhenius base because koh contains

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Answer: hydroxide ions

Explanation:

According to the Arrhenius concept, an acid is a substance that ionizes in the water to give hydronium ion or hydrogen ion and a bases is a substance that ionizes in the water to give hydroxide ion .

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

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2 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago