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nikdorinn [45]
3 years ago
13

Determine the pH of a 0.048 M hypochlorous acid (HClO) solution. Hypochlorous acid is a weak acid (Ka = 4.0 ✕ 10−8 M).

Chemistry
1 answer:
storchak [24]3 years ago
8 0

pH of 0.048 M HClO is 4.35.

<u>Explanation:</u>

HClO is a weak acid and it is dissociated as,

HClO ⇄ H⁺ + ClO⁻

We can write the equilibrium expression as,

Ka = $\frac{[H^{+}] [ClO^{-}]  }{[HClO]}

Ka = 4.0 × 10⁻⁸ M

4.0 × 10⁻⁸ M = $\frac{x \times x }{0.048}

Now we can find x by rewriting the equation as,

x² =  4.0 × 10⁻⁸ × 0.048

   = 1.92 × 10⁻⁹

Taking sqrt on both sides, we will get,

x = [H⁺] = 4.38 × 10⁻⁵

pH = -log₁₀[H⁺]

     = - log₁₀[ 4.38 × 10⁻⁵]

   = 4.35

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snow_tiger [21]
The correct answer is a. This is because the pH of a solution is defined as -log10(concentration of H+ ions). An inverse logarithmic scale such as this means that a solution with a lower concentration of H+ ions will have a higher pH than one with a higher concentration. Therefore we know that the pH of the second sample will be higher than the first.

Since the logarithmic scale has the base 10, a change by 1 on the scale is a consequence of multiplication/division of the H+ concentration by a factor of 10. As the scale is inverse, this means that a decrease of concentration by factor 1000 is equivalent to increasing the pH by (1000/10) = 3.
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How many atoms are in 1.15 mole Cu?
xxMikexx [17]
6.022x10^23 is Avogadro’s number. Use this whenever you work with Stoichiometry involving Atoms, formula units, or molecules. 1 mol of anything is always Avogadro’s number.

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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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The volume increases when the balloon temperature increases.

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-10 F is converted into Kelvin as 249 K.

0°C is nothing but 0+ 273 = 273 K

And the room temperature is 25°C which is converted into Kelvin as 273 + 25 = 298 K.

249 K is below room temperature.

As per the Charles' law volume and temperature are directly proportional to each other, when the pressure of the gas remains constant.

V ∝ T

As the balloon temperature increases, the volume also increases.

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Write in which of these processes heat is absorbed or releases heat.
neonofarm [45]

Answer:

1. Endothermic process

4Exothemic process

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