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alexgriva [62]
2 years ago
6

6)Explain why a 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of

Chemistry
1 answer:
ExtremeBDS [4]2 years ago
5 0

0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of acetic acid, HCH₃COO, is 3.37 is because HCl dissociates more.

<h3>Dissociation</h3>

This is the separation of ions of an ionic compound when it dissolves.

<h3>Hydrogen ion concentration of HCl</h3>

Since 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 its hydrogen ion concentration is gotten from

pH = -log[H⁺]

So,

[H^{+} ] = 10^{-pH}

So, [H⁺] = 10⁻²

<h3>Hydrogen ion concentration of HCH₃COO</h3>

Also, since 0.010 M solution of acetic acid, HCH₃COO, has a pH of 3.37, its hydrogen ion concentration is gotten from

pH = -log[H⁺]

So,

[H^{+} ] = 10^{-pH}

So, [H⁺] = 10⁻³°³⁷ = 0.000427 = 4.27 × 10⁻⁴

Since the hydrogen ion concentration of HCl is 10⁻² and the hydrogen ion concentration of HCH₃COO is 4.27 × 10⁻⁴, we see that HCl dissociates more than HCH₃COO.

So, 0.010 M solution of hydrochloric acid, HCl has a pH of 2.00 but a 0.010 M solution of acetic acid, HCH₃COO, is 3.37 is because HCl dissociates more.

Learn more about dissociation here:

brainly.com/question/25854432

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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