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SCORPION-xisa [38]
3 years ago
6

Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be

en added. Ka of formic acid = 1.8 x 10-4.
Chemistry
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

3.336.

Explanation:

<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>

<em />

So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>

<em></em>

<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>

∵ pH = - log[H⁺].

<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>

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Explanation:

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to make it balanced the reaction should be 3 Mg + N2 = Mg3N2.

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3 years ago
The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compre
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Answer:

0.0126 moles are released

Explanation:

Using Henry's law, where the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid:

S = k×P

<em>Where S is solubility (5.6x10⁻⁴molL⁻¹), k is Henry's constant and P is partial pressure (0.80atm)</em>

Replacing:

<em>5.6x10⁻⁴molL⁻¹ / 0.80atm = 7x10⁻⁴molL⁻¹atm⁻¹</em>

Thus, with Henry's constant, solubility of N₂ when partial pressure is 3.8atm is:

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6.0L × 2.66x10⁻³molL⁻¹ = <em>0.01596 moles of N₂</em>

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Thus, released moles are:

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Answer:

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