Chapter 17.3, Problem 4CP

Calculate the pH after the addition of 35 mL of 0.10 M NaOH to 30 mL of 0.10 M HCN.

a)

11.89

b)

2.11

c)

12.22

d)

1.78

e)

1300

For questions 17.3.5 to 17.3.8, refer to the following diagrams. The solutions shown represent various points in the titration of the weak acid HA with NaOH. (For clarity, the sodium ions and water molecules are not shown.)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution after addition of $\text{NaOH}$ into the $\text{HCN}$ solution is to be determined with given concentration and volume.

Concept introduction:

Acid–base titration is a technique to analyze the unknown concentration of the acid or base with the help of the known concentration of the acid and base.

The equivalence point is the point in an acid–base titration in a chemical reaction where number of moles of the titrant and the unknown concentration of the analyte are equal. It is used to identify the unknown concentration of the analyte.

In strong acid–base titration, the pH of the solution is neutral (i.e., pH = 7) at equivalence point.

In weak acid–strong base titration, the pH of the solution is not neutral (pH> 7) at equivalence point.

In strong acid–weak base titration, the pH of the solution is not neutral (pH < 7) at equivalence point.

The number of moles of the molecule are calculated as:

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}}$.

The number of millimoles of a molecule are calculated as:

$\text{Millimoles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of NaOH}$

For acidic buffer, the value of $\text{pH}$ is calculated using the expression as:

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{Conjugate}\text{Base}\right]}{\left[\text{Weak}\text{Acid}\right]}$.

The $\text{pOH}$ is calculated using the expression:

$\text{pOH}=-\log \left(\left[{\text{OH}}^{-}\right]\right)$

The relationship between $\text{pH}$ and $\text{pOH}$ can be expressed as:

$\text{pH}+\text{pOH}=14$

Answer to Problem 4CP

Solution: Option (a).

Explanation of Solution

Given information:

Theconcentration of $\text{NaOH}$ is $0.10\text{M}$ and the volume is $35\text{mL}$, and theconcentration of $\text{HCN}$ is $0.10\text{M}$ and the volume is $30\text{mL}$.

The equation for reaction between $\text{HCN and NaOH}$ is as follows:

$\text{HCN}\left(aq\right)+\text{NaOH}\left(aq\right)\to \text{NaCN}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

This equation can be simplifiedas follows:

$\text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

After addition of $0.10\text{M}$ concentration of $\text{NaOH}$ and $35\text{mL}$ volume of $\text{NaOH}$ into the $\text{HF}$ solution,

$\text{3}\text{.5}\text{mmoles}$ of $\left[{\text{OH}}^{-}\right]$ neutralize the $3\text{mmoles}$ of $\text{HCN}$, leaving $0.5\text{mmoles}$ of $\text{HF}$ and $0.5\text{mmoles}$ of $\left[{\text{F}}^{-}\right]$ in the solution.

Millimoles of the molecule are calculated using the expression as follows:

$\text{Concentration}=\frac{\text{Milimoles}}{\text{Volume}}$.

Millimoles of $\text{NaOH}$ are calculated as follows:

$\begin{array}{l}\text{Concentration of NaOH}=\frac{\text{Milimoles of NaOH}}{\text{Volume of NaOH}}\\ \text{Milimoles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of NaOH}\end{array}$

Substitute the concentration of $\text{NaOH}$ and the volume of $\text{NaOH}$ in the above expression.

$\begin{array}{l}\text{Millimoles of NaOH}=35\times 0.10\\ =3.5\text{mmoles}\end{array}$

Millimoles of $\text{HF}$ are calculated as follows:

$\begin{array}{l}\text{Concentration of HCN}=\frac{\text{Millimoles of HCN}}{\text{Volume of HCN}}\\ \text{Millimoles of HCN}=\text{Concentration of HCN}\times \text{Volume of HCN}\end{array}$

Substitute the concentration of $\text{HCN}$ and the volume of $\text{HCN}$ in the above expression.

$\begin{array}{l}\text{Millimoles of HCN}=30\times 0.10\\ =3\text{mmoles}\end{array}$

Summarise the millimoles at equilibrium as follows:

$\begin{array}{l}\text{HCN}\left(aq\right)+{\text{ OH}}^{-}\left(aq\right)\rightleftharpoons {\text{H}}_2\text{0}\left(l\right)+\text{ NaCN}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(mmoles\right)& 3& 3.5& -& 0\\ \text{Change}\left(mmoles\right)& 3-3& 3.5-3& -& \text{}3\\ \text{Equilibrium}\left(mmoles\right)& 0& 0.5& -& 3\end{array}\end{array}$

At equilibrium, the solution has astrongbase. The millimoles of $\left[{\text{OH}}^{-}\right]$ are $0.5\text{mmoles}$.

Total volume of the solution is $30+35=65$.

Concentration of $\left[{\text{OH}}^{-}\right]$ is calculated using the expression as follows:

$\text{Concentration}=\frac{\text{Milimoles}}{\text{Total}\text{volume}}$.

Substitute the millimoles of $\left[{\text{OH}}^{-}\right]$ and the total volume of solution in the above expression:

$\begin{array}{l}\text{Concentration}=\frac{0.5\text{mmoles}}{65\text{mL}}\\ =0.00769\text{M}\end{array}$

The $\text{pOH}$ is calculated using the expression:

$\text{pOH}=-\log \left(\left[{\text{OH}}^{-}\right]\right)$

Substitute the concentration of $\left[{\text{OH}}^{-}\right]$ in the above expression:

$\begin{array}{l}\text{pOH}=-\log \left(\left[0.00769\right]\right)\\ =2.11\end{array}$

The $\text{pH}$ is calculated using the expression:

$\text{pH}+\text{pOH}=14$

Substitute the value of $\text{pOH}$ in the above expression:

$\begin{array}{l}\text{pH}=14-\text{pOH}\\ =14-2.11\\ =11.89\end{array}$

The $\text{pH}$ of the solution is 11.89.

Hence, the correct answer is option (a).

Reason for incorrect options:

Option (b) is incorrect because on solving with the help of the above equation, the answer does not match with option (b).

Option (c) is incorrect because on solving with the help of the above equation, the answer does not match with option (c).

Option (d) is incorrect because on solving with the help of the above equation, the answer does not match with option (d).

Option (e) is incorrect because on solving with the help of the above equation, the answer does not match with option (e).

Hence, options(b), (c), (d), and (e) are incorrect.