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antoniya [11.8K]

4 years ago

8

Isopropyl alcohol, (CH3)2CHOH, is a common solvent. Determine the percent by mass of hydrogen in isopropyl alcohol. A) 6.71% H B

) 13.4% H C) 25.0% H D) 53.3% H

1 answer:

vladimir2022 [97]4 years ago

4 0

Answer:

B) 13.4 %.

Explanation:

Isopropyl alcohol has the general molecular formula <em>C₃H₈O.</em>

Its molar mass = 60.1 g/mol.

<em>The percent by mass of H = [(no, of H atoms x molar mass of H) / (molar mass of isopropyl alcohol)] x 100 </em>= [(8.0 atoms x 1.0 g/mol) / (60.1 g/mol)] x 100 = 13.333 % ≅ 13.4 %.

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A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

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$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

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

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

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