Question

What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = 1.8 *10^–5; Kb for C5H5N = 1.7 * 10^–9)

A) 1.0 mole NH3 and 1.5 mole NH4Cl
B) 1.5 mole NH3 and 1.0 mole NH4Cl
C) 1.0 mole C5H5N and 1.5 mole C5H5NHCl
D) 1.5 mole C5H5N and 1.0 mole C5H5NHCl
E) none of these

Answer 1

Answer:

C

Explanation:

When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:

pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])

Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.

a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92

pH = 14 - pOH = 14 - 4.92 = 9.08

b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57

pH = 14 - pOH = 14 - 4.57 = 9.43

c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) =  8.95

pH = 14 - pOH = 14 - 8.95 = 5.05

d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) =  8.59

pH = 14 - pOH = 14 -  = 5.41

Answer 2

A combination of 1.0 mole C₅H₅N and 1.5 mole C₅H₅NHCl will give a buffered solution that has a pH of 5.05.

How to determine the combination of substances required.

In order to determine the combination of substances that is required to produce a buffered solution that has a pH of 5.05, we would apply Henderson-Hasselbalch equations:

$pH =pka+ log_{10} \frac{[A^-]}{[HA]}\\\\pOH = pkb + log_{10} \frac{[A^-]}{[HA]}\\\\pOH = -log_{10}kb + log_{10} \frac{[BH^+]}{[B]}$

Where:

• HA is acetic acid.

• $A^-$  is acetate ion.

Also, the pH of a solution is given by this formula:

pH = 14 - pOH

Note: We used the number of moles to replace the concentration of solutions.

For option A:

• NH₃ = 1.0 mole.

• NH₄Cl = 1.5 mole.

$pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10} \frac{[1.5]}{[1.0]}\\\\pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10}(1.5)$

pOH = 4.92.

pH = 14 - 4.92

pH = 9.08.

For option B:

• NH₃ = 1.5 mole.

• NH₄Cl = 1.0 mole.

$pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10} \frac{[1.0]}{[1.5]}\\\\pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10}(0.6667)$

pOH = 4.57.

pH = 14 - 4.57

pH = 9.43.

For option C:

• C₅H₅N = 1.0 mole.

• C₅H₅NHCl = 1.5 mole.

$pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10} \frac{[1.5]}{[1.0]}\\\\pOH = -log_{10}(1.8 \times 10^{-5}) + log_{10}(1.5)$

pOH = 8.95.

pH = 14 - 8.95

pH = 5.05.

Therefore, a combination of 1.0 mole C₅H₅N and 1.5 mole C₅H₅NHCl will give a buffered solution that has a pH of 5.05.

Read more on moles here: brainly.com/question/3173452