For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge. ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

u i forgot the answer?

it's filtration, but this is an assumption so my apologies if i got this