Q19.55P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the  during the titration of $\mathbf{30}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{0}\mathbf{.}\mathbf{1000}\mathbf{ }\mathbf{\text{M }}\mathbf{ }\mathbf{ }\mathbf{\text{KOH}}$ with $\mathbf{0}\mathbf{.}\mathbf{1000}\mathbf{\text{M}}\mathbf{ }\mathbf{ }\mathbf{\text{HBr}}$ solution after the following additions of acid:

(a) 0 mL (b) 15.00 mL  (c) 29.00 mL (d) 29.90 mL 

(e) 30.00 mL (f) 30.10 mL  (g) 40.00 mL

Step-by-Step Solution

Answer

a) The pH during the titration of $0 \text{mL}$ of $0.1000 \text{KOH}$  with $0.1000\text{ M HBr}$ solution after the $0$ additions of acid is $13$.

b) The pH during the titration of $15.00 \text{mL}$ of $0.1000\text{ KOH}$ with   $0.1000 \text{M HBr}$ solution after the $15.00\text{ mL}$ additions of acid is $12.53$ 

c) The pH during the titration of $29.00 \text{mL}$L of  $0.1000 \text{KOH}$ with $0.1000 \text{M HBr}$ solution after the $15.00\text{ mL}$ additions of acid is $11.23$.

d) The pH during the titration of $29.90 \text{mL}$ of $0.1000 \text{KOH}$ with $0.1000 \text{M HBr}$  solution after the $15.00\text{ mL}$ additions of acid is  $10.23$.

e) The pH during the titration of $30.00 \text{mL}$ of $0.1000 \text{KOH}$ with $0.1000 \text{M HBr}$ solution after the $15.00\text{ mL}$ additions of acid is  $7$.

 f) The pH during the titration of $30.10 \text{mL}$ of $0.1000 \text{KOH}$ with $0.1000 \text{M HBr}$ solution after the  $15.00\text{ mL}$ additions of acid is   $3.77$.

g) The pH during the titration of $40.00 \text{mL}$ of $0.1000 \text{KOH}$ with $0.1000 \text{M HBr}$ solution after the $15.00\text{ mL}$ additions of acid is $1.84$ .

Step 1: Definition of pH

pH is a measure of hydrogen ion concentration, which indicates whether a solution is acidic or alkaline.

Step 2: Subpart (a)

We're titrating a strong base with a strong acid in this situation. To fix this problem, we must first determine whether there is an overabundance of strong acid, in which case we would calculate  $\left[{\text{H}}_3{\text{O}}^{+}\right]$, or is it strong base in the excess, in which case we would calculate  $\left[{\text{OH}}^{-}\right]$ to get pH value. $0.1\text{M} 30\text{ml} \text{KOH}$ and  $0.1\text{M} \text{HBr}$.

We only have a strong base in the solution because no strong acid has been added, thus we use the following equation:

 $$\begin{gathered} \mathrm{pOH}=-\log (0.1\mathrm{M}) \\ =1 \\ \mathrm{pH}=14-\mathrm{pOH} \\ =13. \end{gathered}$$

Therefore, the required pH is 

 $$\begin{gathered} \mathrm{pH}=14-\mathrm{pOH} \\ =14-1 \\ =13 \end{gathered}$$.

Step 3: Subpart (b)

In this example, we add 15mL of strong acid, implying that the strong base is greater than 15mL, and we calculate the concentration of hydroxide ions once more. We must compute in order to answer this challenge. $\left[{\text{OH}}^{-}\right]$ this way:

Total value $:0.03 \text{L}+0.015 \text{L}=0.045 \text{L}$ 

Moles in excess $0.03 \text{L}\times 0.1 \text{M}-0.015 \text{L}\times 0.1 \text{M}=0.0015 \text{mol }$ 

 $$\begin{gathered} \mathrm{pOH}=-\log \frac{0.0015 \text{mol}}{0.045 \text{L}} \\ =1.47 \\ \mathrm{pH}=12.53. \end{gathered}$$

Therefore, the required pH is 12.53.

Step 4: Subpart (c)

In this case we add 29mL of strong acid, which means strong base is in the excess of 1mL and we again calculate concentration of hydroxide ions. In order to solve this problem we have to calculate [OH ] this way: 

Total al value $:0.03\text{ L}+0.029\text{ L}=0.059\text{ L}$

Moles of   in excess: $0.03\text{ L}\times 0.1 \text{M}-0.029\text{ L}\times 0.1 \text{M}=0.0001 \text{mol}$ 

 $$\begin{gathered} \mathrm{pOH}=-\log \frac{0.0001 \text{mol}}{0.059 \text{L}} \\ \mathrm{pOH}=2.77 \\ \mathrm{pH}=14-2.77 \\ =11.23 \end{gathered}$$

Therefore, the required pH is 11.23.

Step 5: Subpart (d)

In this case we add $29.00\text{ mL}$ of strong acid, which means strong base is in the excess of $0.1 \text{mL}$ and we again calculate concentration of hydroxide ions. In order to solve this problem we have to calculate [OH ] this way:

Total value: $0.03 \text{L}+0.0299 \text{L}=0.0599 \text{L}$ 

Moles of   in excess:  $0.03 \text{L}\times 0.1 \text{M}-0.0299 \text{L}\times 0.1 \text{M}=0.00001 \text{mol}$ 

 $$\begin{gathered} \text{pOH}=-\log \frac{0.00001 \text{mol}}{0.0599 \text{L}} \\ =3.77 \\ \text{pH}=10.23. \end{gathered}$$

Therefore, the required pH is 10.23.

Step 6: Subpart (e)

In this case we have same amounts of strong acid and strong base, so in the equivalence point all of the base is neutralised and we know that in the case of total neutralisation of strong base with strong acid the pH is simply equal to 7 .

$\mathrm{pH}=7$

Therefore, the required pH is  7.

Step 7: Subpart (f)

In this case we have the solution after the equivalence point, which means strong acid is in the excess and we can calculate pH by calculating $\left[{\text{H}}_3\text{O}\right]$ in excess:

Total value: $0.03\text{ L}+0.0301\text{ L}=0.0601\text{ L}$ 

Moles of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ in excess : 

 $$\begin{gathered} 0.0301 \text{L}\times 0.1 \text{M}-0.03 \text{L}\times 0.1 \text{M}=0.00001 \text{mol} \\ \text{pH}=-\log \frac{0.00001 \text{mol}}{0.0601\text{ L}} \\ =3.77 \end{gathered}$$

Therefore, the required pH is 3.77.

Step 8: Subpart (g)

In this case we have the solution after the equivalence point, which means strong acid is in the excess and we can calculate pH by calculating $\left[{\text{H}}_3{\text{O}}^{\text{'}}\right]$ in excess:

Total value: $0.03 \text{L}+0.04 \text{L}=0.07 \text{L}$ 

Moles of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ in excess: 

$$\begin{gathered} 0.04\text{ L}\times 0.1 \text{M}-0.03 \text{L}\times 0.1 \text{M}=0.001 \text{mol} \\ \mathrm{pH}=-\log \frac{0.001 \text{mol}}{0.07 \text{L}} \\ =1.84 \end{gathered}$$ 

 

Therefore the required pH is 1.84.