chemistry graduate student is given 450. mL of a 1.10M nitrous acid (HNO,) solution. Nitrous acid is a weak acid with = 4.5 × 10 ". What mass of NaNO, should the student dissolve in the HNO, solution to turn it into a buffer with = 3.18? u may assume that the volume of the solution doesn't change when the NaNO, is dissolved in it. Be sure your answer has unit symbol, and round it to 2 significant digits. Ox10 O
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.
![### Problem Statement: Buffer Solution Preparation
**Objective:**
Calculate the mass of sodium nitrite (NaNO₂) that should be dissolved in a nitrous acid (HNO₂) solution to achieve a desired pH.
**Given Data:**
- Volume of HNO₂ solution: 450 mL
- Initial concentration of HNO₂ solution: 1.10 M
- Acid dissociation constant (Kₐ) for HNO₂: 4.5 × 10⁻⁴
- Desired pH of the buffer solution: 3.18
**Assumptions:**
- The volume of the solution does not change when NaNO₂ is dissolved.
**Steps for Calculation:**
1. **Henderson-Hasselbalch Equation:**
To find the required concentration of the conjugate base (NO₂⁻), we use the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)
\]
where:
- \(\text{pH} = 3.18\)
- \(\text{p}K_a = - \log (4.5 \times 10^{-4}) = 3.35\)
2. **Rearrange the Equation:**
Solving for the ratio of conjugate base (NO₂⁻) to acid (HNO₂):
\[
3.18 = 3.35 + \log \left( \frac{[NO_2^-]}{[HNO_2]} \right)
\]
\[
\log \left( \frac{[NO_2^-]}{[HNO_2]} \right) = 3.18 - 3.35 = -0.17
\]
\[
\frac{[NO_2^-]}{[HNO_2]} = 10^{-0.17} \approx 0.68
\]
3. **Find the Concentration of NO₂⁻:**
\[
[NO_2^-] = 0.68 \times [HNO_2]
\]
Since the initial concentration of HNO₂ is 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38d49af3-6887-43af-8a6c-6e39b942bd2d%2F03c57f78-db24-4b8f-975a-fd2cb8e04877%2Fpvb8u5d_processed.png&w=3840&q=75)

Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images









