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makvit [3.9K]
3 years ago
5

What is the pH of a solution made by mixing 15.00 mL of 0.100 M HCl with 50.00 mL of 0.100 M KOH? Assume that the volumes of the

solutions are additive.
Chemistry
1 answer:
denis23 [38]3 years ago
8 0

Answer:

The correct answer is: pH = 12.73

Explanation:

The <em>neutralization reaction</em> between HCl and KOH is given by the following chemical equation:

HCl + KOH ⇒ KCl + H₂O

Since HCl is a strong acid and KOH is a strong base, HCl is completely dissociated into H⁺ and Cl⁻ ions, whereas KOH is dissociated completely into K⁺ and OH⁻ ions.

For acids, the number of equivalents is given by the moles of H⁺ ions (in this case: 1 equivalent per mol of HCl). For bases, the number of equivalents is given by the moles of OH⁻ ions (in this case: 1 equivalent per mol of KOH).

The H⁺ ions from HCl will react with OH⁻ ions of KOH to give H₂O. The pH is calculated from the difference between the equivalents of H⁺ and OH⁻:

equivalents of H⁺= volume HCl x Molarity HCl

                            = (15.0 mL x 1 L/1000 mL) x 0.100 mol/L

                            = 1.5 x 10⁻³ eq H⁺

equivalents of OH⁻= volume KOH x Molarity KOH

                               = (50.0 mL x 1 L/1000 mL) X 0.100 mol/L

                               = 5 x 10⁻³ eq OH⁻

There are more OH⁻ ions than H⁺ ions. The excess of OH⁻ (that did not react with H⁺ ions) is calculated as follows:

OH⁻ ions= (5 x 10⁻³ eq OH⁻) -  (1.5 x 10⁻³ eq H⁺) = 3.5 x 10⁻³ eq OH⁻= 3.5 x 10⁻³ moles OH⁻  

As the volumes of the solutions are additive, the total volume of the solution is:

V= 15.0 mL + 50.0 mL = 65.0 mL= 0.065 L

So, the concentration of OH⁻ ions in the solution is given by:

[OH⁻] = moles OH⁻/V= (3.5 x 10⁻³ moles OH⁻)/0.065 L = 0.054 mol/L = 0.054 M  

From  [OH⁻], we can calculate pOH:

pOH = -log [OH⁻] = -log (0.054) = 1.27

Finally, we know that pH + pOH= 14; so we calculate pH:

pH= 14 - pOH = 14 - 1,27 =  12.73                                                            

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Answer:

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2) Conversion of fructose 6-phosphate to fructose 1,6-biphosphate by phosphofructokinase

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Explanation:

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A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
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Answer: 32.94 g

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Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

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for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

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Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

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T_{final} = final temperature = 26.6^oC

T_{initial} = initial temperature = 25.0^oC

Now put all the given values in the above formula, we get:

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q=646.4J

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

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