Answer:
Final pH: 9.49.
Round to two decimal places as in the question: 9.5.
Explanation:
The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.
What's the pKb of base B?
Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.
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What's the new salt-to-base ratio?
The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.
Initial:
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After adding the HCl:
;
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Assume that the volume is still 0.5 L:
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What's will be the pH of the solution?
Apply the Henderson-Hasselbalch equation again:
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpOH%7D%20%3D%20%5Ctext%7BpK%7D_b%20%2B%20%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D%20%3D%204.64613%20%2B%20%5Clog%7B%5Cfrac%7B0.760%7D%7B1.04%7D%7D%20%3D%204.50991)
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The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.
There are 7 periods in the modern periodic table
Answer:
percentage of water = 18.76%
Explanation:
A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams.
The overall weight of the compound is 0.4813 grams.
weight of hydrated sample = 0.4813 grams.
weight of anhydrous compound = 0.391 grams
percentage weight of water = mass of water/mass of the hydrated compound × 100
mass of water = mass of hydrated compound - mass of anhydrous compound
mass of water = 0.4813 - 0.391 = 0.0903 grams.
percentage of water = 0.0903/0.4813 × 100
percentage of water = 9.03/0.4813
percentage of water = 18.761687097
percentage of water = 18.76%
Answer:
3. 116.5 V
4. 119.6 V
Explanation:
3. Determination of the voltage.
Resistance (R) = 25 Ω
Current (I) = 4.66 A
Voltage (V) =?
V = IR
V = 4.66 × 25
V = 116.5 V
Thus, the voltage is 116.5 V
4. Determination of the voltage.
Current (I) = 9.80 A
Resistance (R) = 12.2 Ω
Voltage (V) =?
V = IR
V = 9.80 × 12.2
V = 119.6 V
Thus, the voltage is 119.6 V