Chapter 8, Problem 91P

8-101 Suppose you have an aqueous solution prepared by dissolving 0.050 mol of NaH₂PO₄ in 1 L of water. This solution is not a buffer, but suppose you want to make it into one. How many moles of solid Na₂HPO₄ must you add to this aqueous solution to make it into:

(a) A buffer of pH 7.21

(b) A buffer of pH 6.21

(c) A buffer of pH 8.21

Interpretation Introduction

(a)

Interpretation:

The number of moles to be added in NaH₂ PO₄ to make it into a buffer of pH 6.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

$\text{pH }={\text{ pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

This is known as Henderson- Hasselbalch equation. Here, $p{K}_{\text{a}}$ is negative log of acid dissociation constant, $\left[{\text{A}}^{-}\right]$ is concentration of conjugate base and $\left[\text{HA}\right]$ is concentration of acid.

Answer to Problem 91P

$\text{0}{\text{.050 mole of NaH}}_2{\text{PO}}_4$.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH₂ PO₄.

The volume of the solution is 1 L.

The pH of the buffer solution is 7.21.

A buffer is a solution which resists change in pH when limited amounts of an acid or a base are added to it. The given buffer is made of the weak acid, NaH₂ PO₄ and its conjugate base Na₂ HPO₄.

The Henderson-Hasselbalch equation for a buffer of a weak acid (HA) and its conjugate base(A^- ) is given as:

${\text{pH = pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

${\text{pH = pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{\left[{\text{NaH}}_{\text{2}}{\text{PO}}_4\right]}$

As, we know that pK_a of NaH₂ PO₄ is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na₂ HPO₄, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pK_a and the number of moles of NaH₂ PO₄ for [NaH₂ PO₄ ] in the above equation, we get.

$\begin{array}{l}\text{7}\text{.21 = 7}\text{.21}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ 0=\text{log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\end{array}$

Bu taking antilog on both sides we get,

$\begin{array}{l}\text{antilog(0) = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \text{ 1 = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]\text{ = 0}\text{.050}\end{array}$

Thus, $\text{0}\text{.050}$ moles of Na₂ HPO₄ must be added to the solution of NaH₂ PO₄ to make a buffer of 7.21. This also shows that when the equimolar of a weak acid and its conjugate base are taken, the pH of the buffer is equal to the pK_a of the weak acid.

Interpretation Introduction

(b)

Interpretation:

The number of moles to be added in NaH₂ PO₄ to make it into a buffer of pH 7.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

$\text{pH }={\text{ pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

This is known as Henderson- Hasselbalch equation. Here, $p{K}_{\text{a}}$ is negative log of acid dissociation constant, $\left[{\text{A}}^{-}\right]$ is concentration of conjugate base and $\left[\text{HA}\right]$ is concentration of acid.

Answer to Problem 91P

$\text{0}{\text{.005 mole of NaH}}_2{\text{PO}}_4$.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH₂ PO₄.

The volume of the solution is 1 L.

The pH of the buffer solution is 6.21.

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

${\text{pH = pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{\left[{\text{NaH}}_{\text{2}}{\text{PO}}_4\right]}$

As, we know that pK_a of NaH₂ PO₄ is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na₂ HPO₄, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pK_a and the number of moles of NaH₂ PO₄ for [NaH₂ PO₄ ] in the above equation, we get.

$\begin{array}{l}\text{6}\text{.21 = 7}\text{.21}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ -1=\text{log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\end{array}$

Bu taking antilog on both sides we get,

$\begin{array}{l}\text{antilog(-1}\text{.00) = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \text{ 1 = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]\text{ = 0}\text{.005}\end{array}$

Thus, $\text{0}\text{.005}$ moles of Na₂ HPO₄ must be added to the solution of NaH₂ PO₄ to make a buffer of 6.21.

Interpretation Introduction

(c)

Interpretation:

The number of moles to be added in NaH₂ PO₄ to make it into a buffer of pH 8.21 should be calculated.

Concept Introduction:

The pH of buffer solution is calculated using the following formula:

$\text{pH }={\text{ pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

This is known as Henderson- Hasselbalch equation. Here, $p{K}_{\text{a}}$ is negative log of acid dissociation constant, $\left[{\text{A}}^{-}\right]$ is concentration of conjugate base and $\left[\text{HA}\right]$ is concentration of acid.

Answer to Problem 91P

$\text{0}{\text{.500 mole of Na}}_2{\text{HPO}}_4$.

Explanation of Solution

Given Information:

An aqueous solution contains 0.050 moles of NaH₂ PO₄.

The volume of the solution is 1 L.

The pH of the buffer solution is 8.21.

By using the Henderson-Hasselbalch equation for the pH of the buffer solution we get,

${\text{pH = pK}}_{\text{a}}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{\left[{\text{NaH}}_{\text{2}}{\text{PO}}_4\right]}$

As, we know that pK_a of NaH₂ PO₄ is 7.21.

The volume of the buffer solution is 1 L. Assuming that there will not be any change in the volume of solution due to the addition of Na₂ HPO₄, the concentration terms in the Henderson-Hasselbalch equation can be replaced by the number of moles.

By substituting the values of pH, pK_a and the number of moles of NaH₂ PO₄ for [NaH₂ PO₄ ] in the above equation, we get.

$\begin{array}{l}\text{8}\text{.21 = 7}\text{.21}+\text{ log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ 1.00=\text{log}\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\end{array}$

Bu taking antilog on both sides we get,

$\begin{array}{l}\text{antilog(1}\text{.00) = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \text{ 10 = }\frac{\left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]}{0.050}\\ \left[{\text{Na}}_{\text{2}}{\text{HPO}}_4\right]\text{ = 0}\text{.500}\end{array}$

Thus, $\text{0}\text{.500}$ moles of Na₂ HPO₄ must be added to the solution of NaH₂ PO₄ to make a buffer of 8.21.