Determine the [H3O+] of a 0.160M solution of benzoic acid.

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Well, that's quite an acidic question! But don't worry, I won't get sour about it. To determine the [H3O+] of the solution, we need to consider the ionization of benzoic acid, which is a weak acid. When benzoic acid dissolves in water, it partially ionizes, releasing H3O+ ions.

The ionization of benzoic acid can be represented by the equation:

C6H5COOH(aq) + H2O(l) ⇌ C6H5COO-(aq) + H3O+(aq)

Since benzoic acid is a weak acid, we can use the expression for the acid ionization constant (Ka) to find the concentration of H3O+. The Ka for benzoic acid is approximately 6.4 x 10^-5.

Now, to solve for [H3O+], we need to use the equilibrium constant expression:

Ka = [C6H5COO-][H3O+]/[C6H5COOH]

Since we are given the concentration of benzoic acid (0.160 M), we can assume that the amount dissociated is small compared to the initial concentration. Therefore, we can approximate the concentration of the benzoate ion ([C6H5COO-]) to be equal to the concentration of H3O+.

Now, let's plug in all the values:

6.4 x 10^-5 = x^2 / 0.160

Solving this quadratic equation, we find x ≈ 0.0080 M.

Therefore, the [H3O+] of the 0.160 M solution of benzoic acid is approximately 0.0080 M.

Seems like there's a lot of chemistry involved in finding the [H3O+]. But hey, at least we added a little acidity to our conversation!

To determine the [H3O+] of a 0.160M solution of benzoic acid, we need to consider the dissociation of benzoic acid in water. Benzoic acid (C6H5COOH) reacts with water to form hydronium ions (H3O+) and benzoate ions (C6H5COO-), according to the following equation:

C6H5COOH + H2O ⇌ C6H5COO- + H3O+

The balanced equation shows that for every one molecule of benzoic acid that dissociates, one hydronium ion is formed. Therefore, the concentration of hydronium ions is equal to the concentration of benzoic acid.

Hence, the [H3O+] in a 0.160M solution of benzoic acid is 0.160M.

To determine the [H3O+] (concentration of hydronium ions) in a 0.160M solution of benzoic acid, we need to consider the acid's dissociation in water. Benzoic acid, with the chemical formula C6H5COOH, is a weak acid that partially dissociates in water. The dissociation equation for benzoic acid in water is as follows:

C6H5COOH + H2O ⇌ C6H5COO- + H3O+

From this equation, we can see that one mole of benzoic acid reacts with one mole of water to form one mole of benzoate ions (C6H5COO-) and one mole of hydronium ions (H3O+).

Since benzoic acid is a weak acid, it does not fully ionize. Therefore, we can assume that the initial concentration of benzoic acid is also the concentration of hydronium ions formed. In this case, the [H3O+] concentration is 0.160M.

So, the [H3O+] concentration of the 0.160M solution of benzoic acid is 0.160M.

Benzoic acid = HBz

HBz ==> H^+ + Bz^-

Ka = (H^+)(Bz^-)/(HBz)
Set up an ICE chart, substitute into the Ka expression, and solve for (H^+), then convert to pH. You will need to look up the Ka for benzoic acid.