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Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
<u>For the steps below, refer to the ICE chart above.</u>
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know .
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
Imidazole's cyclic structure is aromatic and the lone pair present at the N₁ atom available for donation. hence, according to question N₁ is the most basic atom in the structure.
<h3>What is resonating structure?</h3>- Two π bonds (between C₂-C₃ and N₁-C₅), as well as one lone pair on N₄, can interact with one another to generate a delocalized π system in the cyclic structure.
- This delocalization is intriguing since it has the same number of delocalized electrons as benzene—six.
- As a result, imidazole, like benzene, has a closed, delocalized ring with six π electrons. So, like benzene, it is regarded as an aromatic chemical with resonance stability.
- N₄ is neutral since it cannot be donated because it needs to use its lone pair to be aromatic.
- On the other hand, N₁ already forms a π connection, which helps the system become delocalized.
- N₁ is sp² hybridized and has a trigonal planar basic form. Its lone pair cannot communicate with the delocalized π system since it is pointed away from the cyclic structure.
Know more about Resonating structure
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