Q19.54P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the $\mathit{p}\mathit{H}$ during the titration of $\mathbf{40}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$ of $\mathbf{\text{HCl}}$with  solution after the following additions of base:

(a) $\mathbf{0}\mathbf{ }\mathbf{\text{mL}}$.

(b) $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$.

(c) $\mathbf{39}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$.

(d) $\mathbf{39}\mathbf{.}\mathbf{90}\mathbf{ }\mathbf{\text{mL}}$.

(e) $\mathbf{40}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$.

(f) $\mathbf{40}\mathbf{.}\mathbf{10}\mathbf{ }\mathbf{\text{mL}}$.

(g) $\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{mL}}$.

Step-by-Step Solution

Answer

The pHs during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl }$with a $0.1000\text{M} \text{NaOH}$ solution after the additions of base are:

a) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$$\text{HCl }$with $0.1000\text{M} \text{NaOH}$solution after a 0 ml addition of base is 1.

b) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$$\text{HCl}$with $0.1000\text{M} \text{NaOH}$solution after a 25.00ml addition of base is 1.64.

c) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl}$with $0.1000\text{M} \text{NaOH}$solution after a 39.00ml addition of base is 2.90.

d) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl }$with $0.1000\text{M} \text{NaOH}$solution after a 39.90ml addition of base is 3.90.

e) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl }$with $0.1000\text{M} \text{NaOH}$ solution after a 40.00ml addition of base is 7.

f) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl}$ with $0.1000\text{M} \text{NaOH}$ solution after a 40.10ml addition of base is 10.10.

g) The pH during the titration of $40.00 \text{mL}$of $0.1000\text{M}$ $\text{HCl}$with $0.1000\text{M} \text{NaOH}$ solution after a 50.00ml addition of base is 12.05.

Step 1: Definition of pH

A solution's pH value that measures the concentration of hydrogen ions, reveal whether a solution is acidic or alkaline.

Step 2: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 0   m L addition of base

In this case we have a titration of a strong acid with a strong base. In order to solve this problem, first we have to identify whether the strong acid is in excess or the strong base is in excess. If the former is the case, we would calculate the $\left[{\text{H}}_3{\text{O}}^{+}\right]$, or if the latter is true,we would calculate the $\left[{\text{OH}}^{-}\right]$ to get the pH value. $0.1\text{M} 40\text{ml} \text{HCl}$ and $0.1\text{M} \text{NaCl}$.

a)

In this case we only have a strong acid in the solution, because no strong base is added. So, we will simply apply this equation:

$pH=-\log (0.1M)=1$.

Therefore, the required pH is 1. 

0 . 1000 M   H C l Step 3: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with role="math" localid="1663311793607" 0 . 1000 M   N a O H solution after 25.00 mL the addition of base

In this situation, we add 25 mL of a strong base, indicating that the strong acid’s concentration is greater than 15 mL, and we calculate the concentration of hydronium ions once more. In order to solve this problem, we have to calculate $\left[{\text{H}}_3{\text{O}}^{+}\right]$ the this way:

$$\begin{gathered} Totalvalue=0.04l+0.025l \\ =0.065l. \end{gathered}$$

Mols ${\text{H}}_3{\text{O}}^{+}$of in excess:

$$\begin{gathered} 0.04\text{l}\times 0.1\text{M}-0.025\text{l}\times 0.1\text{M}=0.0015 \text{mol} \\ pH=-\log \frac{0.0015 \text{mol}}{0.065\text{l}} \\ =1.64. \end{gathered}$$

Therefore, the pH is 1.64.

Step 4: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 39.00 mL addition of base

c)

In this case we add 39 ml of a strong base, which means the strong acid is in excess of 1 ml and we again calculate the concentration of hydronium ions. In order to solve this problem, we have to calculate the $\left[{\text{H}}_3{\text{O}}^{+}\right]$ this way:

$$\begin{gathered} Total value=0.04l+0.039 \\ =0.079l. \end{gathered}$$

$$\begin{gathered} Molsof {\text{H}}_3{\text{O}}^{+} inexcess=0.04l\times 0.1M-0.039l\times 0.1M \\ =0.0001 . \end{gathered}$$

$$\begin{gathered} pH=-\log \frac{0.0001mol}{0.079l} \\ =2.90 \end{gathered}$$

Therefore, the pH is 2.90.

Step 5: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 39.90 mL addition o base

d)

In this case we add 39 ml of a strong base, which means the strong acid is in excess of  0.1 ml and we again calculate the concentration of hydronium ions. In order to solve this problem, we have to calculate the $\left[{\text{H}}_3{\text{O}}^{+}\right]$ this way:

$$\begin{gathered} Total value=0.04l+0.0399l \\ =0.0799l. \end{gathered}$$

 

$$\begin{gathered} Mols of {\text{H}}_3{\text{O}}^{+} in excess=0.04l\times 0.1M-0.0399l\times 0.1M \\ =0.00001 . \end{gathered}$$mol.

$$\begin{gathered} pH=-\log \frac{0.00001 \text{mol}}{0.0799l} \\ =3.90. \end{gathered}$$.

Therefore, the pH is 3.90.

Step 6: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 40.00 mL addition of base

e)

We have an identical amount of a strong acid and a strong base, therefore, in this situation, at the equivalence point, all of the acid is neutralized, and we know that the pH is simply equal to 7 in the event of a total neutralization of a strong acid with a strong base.

Therefore, the pH is 7 .

Step 7: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 40.10 mL addition of base

f)

We get the answer after the equivalence point in this case, which means a strong base is in excess. We can calculate the pH by calculating the  in excess:

$$\begin{gathered} Totalvalue=0.04l+0.0401 l \\ =0.0801 l. \end{gathered}$$

Moles of $\text{OH'}$in excess:

 $$\begin{gathered} \text{0}.\text{0401l}\times \text{0}.\text{1M - 0}.\text{04l}\times \text{0}.\text{1M = 0}.\text{00001} \text{mol} \\ pOH=-\log \frac{0.00001 \text{mol}}{0.0801l} \\ =3.9. \\ pH=14-pOH \end{gathered}$$

Therefore, the pH is 10.10.

Step 8: Calculate the pH of 40.00 mL of 0 . 1000 M   H C l with 0 . 1000 M   N a O H solution after the 50.00 mL addition of base

g)

We get the answer after the equivalence point in this case, which means a strong base is in excess. We can calculate the pH by calculating the ${\text{OH}}^{-}$in excess:

$$\begin{gathered} Total value=0.04l+0.05l \\ =0.09l \end{gathered}$$

Moles $\text{OH - }$of in excess:

$$\begin{gathered} 0.05l\times 0.1M-0.04l\times 0.1M=0.001 \text{mol} \\ pOH=-\log \frac{0.001 \text{mol}}{0.09l} \\ =1.95. \\ pH=14-pOH \\ =12.05. \end{gathered}$$

Therefore, the required pH is 12.05 .