Answer:
The answers are in the explanation
Explanation:
- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.
- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.
- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.
- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.
I hope it helps!
I'm not sure what you mean. Besides, I feel like you're talking math.
But anyways, if you have 120 let's say
The scientific notation is 1.20 × 10^2
if you have 125000
the scientific notation is 1.25 × 10^ 5
The number of times you go left the decimal, I guess exponent increases
So yea
Answer : The value of Ka for acetic acid is, 
Explanation :
The chemical formula of acetic acid is, 
.
The chemical equilibrium reaction will be:
Given:
pH = 2.96
First we have to calculate the concentration of hydrogen ion.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![2.96=-\log [H^+]](https://tex.z-dn.net/?f=2.96%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=1.096\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.096%5Ctimes%2010%5E%7B-3%7DM)
That means,
![[H^+]=[CH_3COO^-]=1.096\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BCH_3COO%5E-%5D%3D1.096%5Ctimes%2010%5E%7B-3%7DM)
![[CH_3COOH]=0.0602-(1.096\times 10^{-3})=0.0591M](https://tex.z-dn.net/?f=%5BCH_3COOH%5D%3D0.0602-%281.096%5Ctimes%2010%5E%7B-3%7D%29%3D0.0591M)
The expression for reaction is:
![K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BCH_3COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3COOH%5D%7D)
Thus, the value of Ka for acetic acid is,
