Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
A solution of sodium cyanide, NaCN, has a pH of 11.20. How many grams of NaCN are in 835 mL of a solution with the same pH?
![**Educational Lab Exercise: Calculating the Mass of Sodium Cyanide in Solution**
---
### Problem Statement
A solution of sodium cyanide, NaCN, has a pH of 11.20. How many grams of NaCN are in 835 mL of a solution with the same pH?
Given:
- \( K_b \) (CN\(^-\)) = \( 1.7 \times 10^{-5} \)
**Mass of NaCN (g):**
\[ \boxed{ \quad \quad } \]
---
### Steps to Solve:
1. **pH to pOH Conversion**: Given the pH, first find the pOH.
\[
\text{pOH} = 14 - \text{pH}
\]
Substitute pH = 11.20:
\[
\text{pOH} = 14 - 11.20 = 2.80
\]
2. **Hydroxide Ion Concentration**: Calculate the concentration of OH\(^-\) ions:
\[
\left[ \text{OH}^- \right] = 10^{-\text{pOH}} = 10^{-2.80}
\]
3. **Henderson-Hasselbalch Equation**:
Using the equilibrium expression of the weak base:
\[
K_b = \frac{\left[ \text{OH}^- \right] \left[ \text{HCN} \right]}{\left[ \text{CN}^- \right]}
\]
Since the problem involves the complete dissociation of NaCN:
\[
\left[ \text{CN}^- \right] = \left[ \text{OH}^- \right]
\]
4. **Equilibrium Concentration**:
\(
\left[ \text{OH}^- \right] = \sqrt{ K_b \times \text{initial concentration of NaCN} }
\)
Plugging in the known values will let us find the initial concentration of NaCN, and subsequently using:
\[
\text{initial concentration of NaCN} (\text{mol/L}) = \frac{\text{mass of NaCN}}{ \text{molar mass of NaCN} } \times \frac{1}{\text{Volume in Liter}}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb91a8d0c-ccba-43d9-b564-e4e05a79bb72%2F65656eda-1b4c-410c-8519-a3fb9334191f%2Ffwbuwei_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
