Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have
= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is
Number of moles of 
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of 
= 0.167 M
Now the ICE table :
I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,
= 
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)
So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= 
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
When water vapor in the atmosphere loses heat and cools down, condensation happens. As the water vapor cools down and condenses, it attaches to small particles of dust floating in the atmosphere, forming tiny liquid water droplets.
Like mitosis, meiosis is a form of eukaryotic cell division. However, these two processes distribute genetic material among the resulting daughter cells in very different ways. meiosis gives rise to four unique daughter cells, each of which has half the number of chromosomes as the parent cell. Because meiosis creates cells that are destined to become gametes (or reproductive cells), this reduction in chromosome number is critical — without it, the union of two gametes during fertilization would result in offspring with twice the normal number of chromosomes!
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL
