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Nataly [62]
3 years ago
6

What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml

of solution
Chemistry
1 answer:
Dimas [21]3 years ago
7 0

Answer:

pH= 12

Explanation:

Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:

KOH⇒ K⁺ + OH⁻

Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:

[OH⁻]= [KOH]

In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):

Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol

moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles

The molality of the solution is the number of moles of KOH per liter of solution:

V= 250.0 ml x 1 L/1000 ml= 0.250 L

M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M

Now, we calculate pOH:

pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2

Finally, we calculate pH from pOH:

pH + pOH = 14

⇒pH = 14 - pOH= 14 -2 = 12

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Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

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If you want to study two different states, an initial state 1 and a final state 2, the following is true:

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<h3>Temperature of the gas in this case</h3>

In this case, you know:

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Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

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The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

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Explanation:

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