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GarryVolchara [31]

3 years ago

5

A 25.0 mL sample of a saturated C a ( O H ) 2 solution is titrated with 0.029 M H C l , and the equivalence point is reached aft

er 37.3 mL of titrant are dispensed. Based on this data, what is the concentration (M) of the hydroxide ion?

1 answer:

Gnesinka [82]3 years ago

3 0

Answer:

0.043 M

Explanation:

The reaction that takes place is:

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume required to reach the equivalence point</em>:

0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺

As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.

With that in mind we can <u>calculate the hydroxide ion concentration in the original sample solution</u>, using <em>the calculated number of moles and given volume</em>:

1.0817 mmol OH⁻ / 25.0 mL = 0.043 M

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Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends

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24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.

$E_\text{k} = 1/2 \; m \cdot v^{2}$

Where

$E_\text{k}$ is the <em>kinetic energy</em> of the molecule,
$m$ its mass, and
$v^{2}$ the square of its speed.

Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,

$E_\text{k} \propto T$

where $\text{T}$ its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃

$E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)$

$m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

Where

$m(\text{HCl})$ , $v(\text{HCl})$ , and $E_\text{k}(\text{NH_3})$ the mass, speed, and kinetic energy of an HCl molecule;
$m(\text{NH}_3)$ , $v(\text{NH}_3)$ , and $E_\text{k}(\text{NH}_3)$ the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, $m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3)$ . Therefore:

$36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

$36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)$

$\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)$

The <em>average </em>speed NH₃ molecules would be $\sqrt{36.45/17.03} \approx 1.463$ <em>if</em> the <em>average </em>speed of HCl molecules $v(\text{HCl})$ is 1.

$\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}$

$\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}$

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3 years ago