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3 years ago

What is the volume of a salt crystal measuring 5.44 x 10 ^ - 2 m by 2.5 x 10 ^ - 3 m by 7.9 x10-^ 3 m ?

Chemistry

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

The volume of a salt crystal with the mentioned dimensions is V= 107.44 x 10^-8.

Explanation:

To solve this problem we multiply the length of the sides of the crystal. To calculate this product you have to calculate:

1) (5.44 x 2.5 x 9,9). and

2) (10^-2 x 10^-3 x 10^-3). Here we have the product of powers of equal base so we should sum the exponents.

⇒(10^-2 x 10^-3 x 10^-3) = 10^(-2 + -3 + -3) ⇒ (10^-2 x 10^-3 x 10^-3) = 10^-8.

Finally,

V= (5.44 x 10^-2)x (2.3 x 10^-3) x (7.9 x 10^-3) ⇒ V= 107.44 x 10^-8 m.

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Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

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$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

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$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

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