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Charra [1.4K]
2 years ago
13

Calculate the concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58.

Chemistry
1 answer:
Artyom0805 [142]2 years ago
8 0

The concentration (in M) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

pOH is the measure of basic nature of a solution by evaluating the [OH⁻] concentration.

pH is measure of acidic nature of a solution by evaluating the [H⁺] concentration.  It is the negative logarithm of the hydroxide ion concentration. It gives hydronium ions on dissociation.

Given,

pOH = 3.58

Temperature = 25°C = 298K

We know that pH + pOH =14

pH + 3.58 = 14

pH = 10.42

pH is the negative logarithm of hydrogen ion concentration.

At 25°C, the relation of pH and [H⁺] concentration is as follows:

∴ pH = -log [H⁺]

⇒ 10.42 = -log [H⁺]

⇒log [H⁺] = -10.42

⇒ [H⁺] = antilog (-10.42)

⇒[H⁺] =3.8× 10⁻¹¹ M

The concentration (in m) of hydronium ions in a solution at 25. 0 °C with a POH of 3. 58 Is 3.8× 10⁻¹¹ M

Learn more about pH here, brainly.com/question/17144456

#SPJ4

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Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

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Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

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