Picric acid, or 2,4,6 -trinitrophenol, has the following...

Picric acid, or 2,4,6 -trinitrophenol, has the following structure: Not only is it an acid, it is also an explosive. (a) Based on what you know of oxyacids, which hydrogen ionizes from the picric acid molecule? (b) Picric acid has a $K_{\mathrm{a}}$ of $0.380 .$ What is the $\mathrm{pH}$ of a $0.100 M$ solution of picric acid? You will have to use the quadratic formula. (c) $\Delta$ When picric acid explodes anaerobically (that is, without oxygen), it forms carbon monoxide, water, nitrogen, and solid carbon. Write the balanced chemical equation for the decomposition of picric acid. (d) $\Delta$ Would substitution of the nitro $\left(\mathrm{NO}_{2}\right)$ groups in the molecule with amino $\left(\mathrm{NH}_{2}\right)$ likely increase or decrease the $K_{\mathrm{a}}$ of the compound? Explain your answer.

Picric acid, or 2,4,6 -trinitrophenol, has the following structure:
Not only is it an acid, it is also an explosive.
(a) Based on what you know of oxyacids, which hydrogen ionizes from the picric acid molecule?
(b) Picric acid has a $K_{\mathrm{a}}$ of $0.380 .$ What is the $\mathrm{pH}$ of a $0.100 M$ solution of picric acid? You will have to use the quadratic formula.
(c) $\Delta$ When picric acid explodes anaerobically (that is, without oxygen), it forms carbon monoxide, water, nitrogen, and solid carbon. Write the balanced chemical equation for the decomposition of picric acid.
(d) $\Delta$ Would substitution of the nitro $\left(\mathrm{NO}_{2}\right)$ groups in the molecule with amino $\left(\mathrm{NH}_{2}\right)$ likely increase or decrease the $K_{\mathrm{a}}$ of the compound? Explain your answer.

Key Concepts

Oxyacids

Oxyacids are acids that contain hydrogen, oxygen, and another element, typically with the acidic hydrogen bonded to an oxygen atom. In these acids, the hydrogen ion (proton) that is released during ionization usually comes from a hydroxyl group attached to the central atom. Understanding which hydrogen ionizes involves analyzing the molecular structure and resonance stabilization effects that influence acid strength.

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant, Ka, quantitatively expresses the extent to which an acid dissociates into its ions in solution. A higher Ka value indicates a stronger acid and vice versa. The negative logarithm of Ka, known as the pKa, provides a more convenient scale for comparing acid strengths. This parameter is central to predicting the behavior of weak acids in solution.

Weak Acid Equilibrium Calculations

For weak acids that do not fully dissociate in solution, equilibrium calculations are necessary to determine the concentrations of all species in solution. When the equilibrium concentrations are not easily approximated, the quadratic formula can be applied to solve for the concentration of hydrogen ions, making such calculations essential for accurately determining the pH of the solution.

Balancing Chemical Equations

Balancing chemical equations is a fundamental aspect of chemistry that ensures the law of conservation of mass is obeyed in chemical reactions. In complex reactions, such as decomposition reactions that produce multiple products, it is important to correctly balance all atoms on both sides of the reaction, accounting for different phases and states of matter.

Substituent Effects on Acidity

Substituent effects refer to the influence that different chemical groups attached to a molecule have on its acidity. Electron-withdrawing groups stabilize the negative charge on the conjugate base, thereby increasing acid strength, while electron-donating groups destabilize it, decreasing acidity. Understanding these effects helps predict changes in the acid dissociation constant when the substituents in a molecule are altered.

Transcript

00:01 Now looking at the structure of our compound for the first part of this question, we are looking at n positive or a double bond to the oxygen.

00:09 We have a ring structure and right here we've got the o and we've got the h and here we've got the n positive double bond all to another o.

00:25 We've got an h, we've got an n slightly positive or other positive a double bond and then we've got an o and then finally we've got an h.

00:34 Now looking at the ionization process we are looking at the hydrogen which is bonded to the oxygen atom.

00:42 So this is the hydrogen that is going to be ionized from the bulk of this molecule, leaving this with a negative charge.

00:52 And then as for the next part of the question, what we are looking at here is hc6, h5, m3 or 7, ionizing in h2 or partially to form h3o positive and say 6 h2 and 3 or 7 negative let's just call this a and let's just call this a negative for simplicity now if we look at the initial concentrations we have set up 1 and before the ionization takes place we won't be having any of these in solution then looking at the change in the concentrations due to the ionization process.

01:33 If this decreases by x, it means this will increase by x and this will increase by x since they are in the ratios of 1 is to 1.

01:41 So at equilibrium we've got 0 .1 minus x of a, we've got x of our h3 or positive and we've got x of the a negative ion.

01:51 Now if we use the ka expression, we're saying ka is equal to the concentration of a negative multiplied by the concentration of h3 or positive divided by the concentration of a...

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