If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
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What is base dissociation constant?
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The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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PH is defined as the negative log of Hydronium Ion concentration.
So, in order to find the pH of vinegar, we find the negative log of its hydronium ion concentration.
![pH=-log[ H_{3}O^{+}] \\ \\ pH=-log(1.6 * 10^{-3}) \\ \\ pH=2.8](https://tex.z-dn.net/?f=pH%3D-log%5B%20H_%7B3%7DO%5E%7B%2B%7D%5D%20%5C%5C%20%20%5C%5C%20%0ApH%3D-log%281.6%20%20%2A%2010%5E%7B-3%7D%29%20%20%5C%5C%20%20%5C%5C%20%0ApH%3D2.8)
So, the pH of given vinegar solution will be 2.8.
Therefore, the answer to this question is option B
The answer for this would be 69.6
Bonding Continuum<span>. Trends in the Periodic Table and </span>Bonding<span>. Ionic </span>Bonds. Ionic bonds<span> are formed between atoms with a large difference in electronegativities. ... The ionic </span>bond<span> is the electrostatic force of attraction between a positive and negative ion.</span>
Answer:
The molarity of the dissolved NaCl is 6.93 M
Explanation:
Step 1: Data given
Mass of NaCl = 100.0 grams
Volume of water = 100.0 mL = 0.1 L
Remaining mass NaCl = 59.5 grams
Molar mass NaCl= 58.44 g/mol
Step 2: Calculate the dissolved mass of NaCl
100 - 59. 5 = 40.5 grams
Step 3: Calculate moles
Moles NaCl = 40.5 grams / 58.44 g/mol
Moles NaCl = 0.693 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity dissolved NaCl = 0.693 moles / 0.1 L
Molarity dissolved NaCl = 6.93 M
The molarity of the dissolved NaCl is 6.93 M
