How many grams of pyridine are there in 100 mL of an aqueous solution that has a pH of 9.00? The K_b for pyridine is 1.9 × 10^-9 and the equation of interest is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq). A)0.053 g B)0.42 g C)0.79 g D)7.9 g

Question

Quizplus · Accepted Answer

First, we need to use the given equation and the Ka value to find the concentration of pyridine in the solution.

Ka = [C5H5NHS+][OH-]/[C5H5N]

At pH 9, [OH-] = 1.0 x 10^-5 M

Let x be the concentration of pyridine in M

1.9 x 10^-9 = (x^2)/(x + 1.0 x 10^-5)

Solving for x, we get x = 1.22 x 10^-4 M

Now, we can use the density of water (1 g/mL) and the molar mass of pyridine (79.1 g/mol) to convert the concentration to grams per mL and then multiply by 100 mL to get the total grams of pyridine:

1.22 x 10^-4 M * 79.1 g/mol * 1 mL/1 g = 0.00965 g/mL

0.00965 g/mL * 100 mL = 0.965 g

However, we need to account for the fact that pyridine is not the only component of the solution - it is dissolved in water and there may be some pyridine hydroxide present. To do this, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([C5H5NHS+]/[C5H5N])

At pH 9, we can assume that [C5H5N] is much larger than [C5H5NHS+], so we can simplify:

9.00 = 8.72 + log([C5H5NHS+]/[C5H5N])

0.28 = log([C5H5NHS+]/[C5H5N])

[C5H5NHS+]/[C5H5N] = 1.93

This means that for every 1.93 moles of pyridine, there is 1 mole of pyridine hydroxide. So we can adjust our concentration of pyridine accordingly:

1.22 x 10^-4 M / 2.93 = 4.15 x 10^-5 M

And now we can calculate the grams of pyridine:

4.15 x 10^-5 M * 79.1 g/mol * 1 mL/1 g = 0.00329 g/mL

0.00329 g/mL * 100 mL = 0.329 g

Therefore, the answer is closest to B) 0.42 g.

How Many Grams of Pyridine Are There in 100 ML