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3 years ago

Which of these solutions is acidic?

Chemistry

1 answer:

padilas [110]3 years ago

6 0

Answer:

Option a is a solution acidic ([OH⁻] = 7.0x10⁻⁹ M).

Explanation:

To know if a solution is acidic we need to calculate the pH and it must be lower than 7. A value of pH equal to 7 is a neutral solution and a solution with a pH value higher than 7 is a basic solution.

a. For the [OH⁻] = 7.0x10⁻⁹ M we have:

$pOH = -log[OH^{-}] = -log(7.0 \cdot 10^{-9}) = 8.15$

Now, the pH is:

$pH + pOH = 14$

$pH = 14 - pOH = 14 - 8.15 = 5.85$

This solution is acidic (pH < 7)

b. [H₃O⁺] = 8.5x10⁻⁸ M

$pH = -log(8.5 \cdot 10^{-8}) = 7.07$

This is not an acidic solution. Is a neutral one (pH around to 7).

c. [OH⁻] = 2.5x10⁻⁶ M

$pOH = -log[OH^{-}] = -log(2.5 \cdot 10^{-6}) = 5.60$

Then, the pH is:

$pH = 14 - pOH = 14 - 5.60 = 8.40$

Hence, this is not an acidic solution. It is basic (pH > 7).

Therefore, option a is a solution acidic ([OH⁻] = 7.0x10⁻⁹ M).

I hope it helps you!

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Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

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$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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