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2 years ago

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF

Chemistry

1 answer:

stealth61 [152]2 years ago

4 0

Answer:

247 ml

Explanation:

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml

0.000150 x ? = 0.0370 moles HF

? = 0.0370/0.000150 = 247 ml

check

247 ml = 247/1000 L = 0.247

(0.247) x (0.150) =0.370 check

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Read 2 more answers

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

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