Explanation:
The given reaction is as follows.

Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,

Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] = 
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that
of the weak acid is 3.42.
Answer:
2%
Explanation:
oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.
1 significant figure, because there is no decimal after the zero the zero doesn't count.
2. Rubidium
3. Antimony
4. Ytterbium
5. Einsteinium
Answer: Do want the experiment done or help of the experiment