Question

One liter of a 0.1M Tris buffer (pKa=8.3) is adjusted to a pH of 2.0. a. What are the concentrations of the conjugate base and weak acid at this pH? b. What is the pH when 1.5mL of 3.0M HCl is added to this buffer? Is Tris a good buffer at this pH? Why? c. What is the pH when 1.5mL of 3.0M NaOH is added to this buffer?

Answer 1

Answer:

a. [Weak acid] = 0.099999949M

[Conjugate base = 5.0119x10⁻⁸M

b.  pH = 1.84. Is not a good buffer

c. pH = 6.97

Explanation:

a. To find pH of the buffer we must use Henderson-Hasselbalch formula:

pH = pKa + log [Conjugate base] / [Weak acid]

2.0 = 8.3 + log [Conjugate base] / [Weak acid]

5.01x10⁻⁷ = [Conjugate base] / [Weak acid] (1)

As concentration of the buffer is 0.1M = [Conjugate base] + [Weak acid] (2)

Replacing (2) in (1):

5.01x10⁻⁷ = 0.1M - [Weak acid] / [Weak acid]

5.01x10⁻⁷ [Weak acid] = 0.1M - [Weak acid]

[Weak acid] = 0.099999949M

[Conjugate base] = 5.0119x10⁻⁸M

b. The conjugate base reacts with the HCl. Moles of HCl are:

1.5x10⁻³L * (3.0mol / L) = 4.5x10⁻³ moles HCl

As the conjugate base contains just 5x10⁻⁸ moles. Almost all HCl doesn't react and pH =

[H⁺] = 4.5x10⁻³ moles + 1x10⁻² moles (Initial moles H⁺) / L = 0.0145M

pH = -log [H⁺]

pH = 1.84

As the concentration of the conjugate base is <<< than weak acid. In this pH, Tris is not a good buffer. As general rule a good buffer works pH between pKa ± 1.

c. Now, NaOH reacts with the weak acid producing conjugate base.

The new moles are:

[Weak acid] = [Weak acid] = 0.099999949M - 4.5x10⁻³ = 0.0955

[Conjugate base] = 5.0119x10⁻⁸M + 4.5x10⁻³ = 4.5x10⁻³

pH = pKa + log [Conjugate base] / [Weak acid]

pH = 8.3 + log [4.5x10⁻³] / [0.0955]

pH = 6.97

Answer 2

Following are the response to the given points:

For question a)

The concentration of Tris buffer $\text{(Tris buffer)} = 0.1\ M$

Using formula:

$\text{[ Tris buffer ] = [ Conjugate base ] + [ Acid ]}$

$0.1 M = \text{ Conjugate base +Acid } \\\\ \text{Acid} = 0.1\ M - \text{ Conjugate base } .............(a)$

Using the Henderson-Hasselbalch equations

$\to pH = pKa + \log \frac{\text{ Conjugate base }}{ Acid}\\\\\to 2.0 = 8.3 + \log \frac{\text{ Conjugate base }}{ Acid}\\\\\to \log \frac{\text{ Conjugate base }}{ Acid} = 2.0 - 8.3\\\\\to \log \frac{\text{ Conjugate base }}{ Acid} = - 6.3\\\\\to \frac{\text{ Conjugate base }}{ Acid}= 10-6.3\\\\\to \frac{\text{ Conjugate base }}{ Acid} = 5.01 \times 10^{-7}\\\\\to \text{Conjugate base} = 5.01 \times 10^{-7} \times \ Acid ..............(b)$

Putting the value of equation (b) in equation (a):

$\text{Acid } = 0.1\ M - 5.01 \times 10^{-7}\\\\\text{Acid } = 0.1 M \ ( approx.)$

from equation (b)

$\to \text{Conjugate base} = 5.01 \times 10^{-7} \times 0.1\ M= 5.01 \times 10^{-8}\ M$

Therefore,

$\to \text{Acid} = 0.1\ M\\\\ \to \text{Conjugate base} = 5.01 \times 10^{-8}\ M$

For question b)

$\to \text{Number of moles of HCl} = 3.0\ M \times 0.00152\ L= 0.0045 \ moles\\\\\to \text{Number of moles Conjugate base in 1 L of buffer} = 5.01 \times 10^{-8}\ M \times 1\ L = 5.01 \times 10^{-8}\ moles\\\\ \to \text{Number of moles of weak acid} = 0.1\ M \times 1\ L = 0.1\ moles$

Now we need all the conjugate bases for neutralizing the $HCl$

therefore

Calculating the molarity of  $HCl$ in the solution $= \frac{0.0045\ moles}{1\ L} = 0.0045\ M$

$\to H^+ \ in \ HCl = 0.0045 \ M\\\\p_H = 2.0\\\\\therefore \\\\ H^+ \ in \ buffer = 10^{-2}\ M = 0.01\ M \ ( \because \ p_H = - \log [ H^+]\ and\ [ H^+ ] = 10-p_H )$

Now Total concentration of $H^+ \ i.e\ [H^+]$

$\to Total = 0.0045 + 0.01 = 0.0145\ M\\\\\to pH = - \log 0.0145\\\\\to pH = - ( - 1.84 )\\\\\to pH = 1.84$

For question c)

Number of moles of $NaOH = 3.0 \ M \times 0.00152\ L= 0.0045\ moles$

$\to \text{Number of moles Conjugate base ( say A-) in 1\ L of buffer} = 5.01 \times 10^{-8}\ M \times 1 \ L = 5.01 \times 10^{-8}\ moles \\\\ \to \text{Number of moles of weak acid ( say HA )} = 0.1\ M \times 1\ L = 0.1\ moles$

Now ICF table is :

$HA + OH^- \longrightarrow A^- + H_2O$

$\text{I 0.1 mol 0.0045 mol 5.01 x 10-8 mol}\\\\\text{C - 0.0045 mol - 0.0045 mol + 0.0045 mol}\\\\\text{F 0.0955 mol 0 mol 0.0045 mol ( approx. )}$

Using Henderson-Hasselbalch:

$\to pH = 8.3 + \log \frac{0.0045}{ 0.0955} \\\\\to pH = 8.3 + \log 0.04712\\\\\to pH = 8.3 - 1.3268\\\\\to pH = 6.97$

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