gmail Google Docs Netflix Disney+ spotify Hulu Amazon Homepage outlook mail Green and White H... Blackboard Courses Acids and Bases Calculating the pH of a salt solution Calculate the pH at $$ 25^{\circ} \mathrm{C} $$ of a 0.62 M solution of lidocaine $$ \mathrm{HCl}\left(\mathrm{C}_{14} \mathrm{H}_{21} \mathrm{NONH}_{2} \mathrm{Cl}\right) $$. Note that lidocaine $$ \left(\mathrm{C}_{14} \mathrm{H}_{21} \mathrm{NONH}\right) $$ is a weak base with a $$ p K_{b} $$ of 7.94 . Round your answer to 1 decimal place.

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1. We are given that lidocaine is a weak base with  
   $$
   pK_b = 7.94.
   $$
   Its conjugate acid (lidocaine HCl) will have a $$ pK_a $$ given by
   $$
   pK_a = 14 - pK_b = 14 - 7.94 = 6.06.
   $$
   
2. The acid dissociation constant is
   $$
   K_a = 10^{-pK_a} = 10^{-6.06}.
   $$
   
3. The salt (lidocaine HCl) dissociates completely in water:
   $$
   \text{HA} \rightarrow \text{H}^+ + \text{A}^-,
   $$
   where HA represents the conjugate acid of lidocaine. The initial concentration of HA is 0.62 M.

4. Let $$ x $$ be the concentration of $$\text{H}^+$$ formed from the weak acid dissociation:
   $$
   \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-.
   $$
   Then, at equilibrium, we have:
   $$
   [\text{H}^+] = x,\quad [\text{A}^-] = x,\quad [\text{HA}] = 0.62 - x.
   $$

5. The equilibrium expression is
   $$
   K_a = \frac{x^2}{0.62 - x}.
   $$
   Because $$ K_a $$ is very small, we can assume $$ x \ll 0.62 $$, and approximate:
   $$
   0.62 - x \approx 0.62.
   $$
   Thus,
   $$
   x^2 \approx K_a \times 0.62.
   $$

6. Substitute $$ K_a $$ into the expression:
   $$
   x \approx \sqrt{10^{-6.06} \times 0.62}.
   $$
   With
   $$
   10^{-6.06} \approx 8.7 \times 10^{-7},
   $$
   we have:
   $$
   x \approx \sqrt{(8.7 \times 10^{-7})(0.62)}.
   $$

7. Compute the product:
   $$
   (8.7 \times 10^{-7})(0.62) \approx 5.394 \times 10^{-7}.
   $$
   Taking the square root:
   $$
   x \approx \sqrt{5.394 \times 10^{-7}} \approx 7.35 \times 10^{-4} \text{ M}.
   $$

8. The pH is given by:
   $$
   \text{pH} = -\log (x) = -\log (7.35 \times 10^{-4}).
   $$
   Noting that
   $$
   \log (7.35 \times 10^{-4}) = \log (7.35) - 4 \approx 0.866 - 4 = -3.134,
   $$
   we obtain:
   $$
   \text{pH} \approx 3.13.
   $$

9. Rounding to 1 decimal place, the pH is:
   $$
   \boxed{3.1}.
   $$
The pH of the 0.62 M lidocaine HCl solution at $$ 25^{\circ} \mathrm{C} $$ is **3.1**.

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