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.
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}](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)
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![\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BpK%7D_b%20%3D%20%5Ctext%7BpOH%7D%20-%5Clog%7B%5Cfrac%7B%5B%5Ctext%7BSalt%7D%5D%7D%7B%5B%5Ctext%7BBase%7D%5D%7D%7D%5C%5C%5Cphantom%7B%5Ctext%7BpK%7D_b%7D%20%3D%204.5%20-%20%5Clog%7B%5Cfrac%7B0.750%7D%7B1.05%7D%7D%20%5C%5C%5Cphantom%7B%5Ctext%7BpK%7D_b%7D%20%3D4.64613)
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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:
![\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BB%7D%5D%20%3D%20%5Cfrac%7Bn%7D%7BV%7D%20%3D%20%5Cfrac%7B0.520%7D%7B0.5%7D%20%3D%201.04%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.![\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BBH%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7Bn%7D%7BV%7D%20%3D%20%5Cfrac%7B0.380%7D%7B0.5%7D%20%3D%200.760%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
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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.
Vapor pressure is a criteria for a substance's volatility. This is the ability of a substance to transition from liquid to vapor. When the vapor pressure equals the external pressure, liquid turns to gas. Hence, if the substance has higher vapor pressure, then it is volatile. <em>Therefore, Sample A is more volatile than Sample B.</em>
Answer:
11.7 mL
Explanation:
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question ,
The density of the metal = 10.5 g/cm³
The mass of the metal = 5.25 g
Hence , the volume can be calculated from the above formula , i.e. ,
d = m / V
V = m / d
V = 5.25 g / 10.5 g/cm³
V = 0.5 cm³
Since , The unit 1 mL = 1 cm³
V = 0.5 mL
The vessel has 11.2 mL of water ,
The new volume becomes ,
11.2 mL + 0.5 mL = 11.7 mL
Answer:
True
Explanation:
The physical changes are reversible in most cases and these changes are not the chemical changes which means that it is only the change in its state not in their nature. Just take the example of water, on cooling it becomes solid and change in color can be seen which is white in solid form and colorless in liquid form. This is also reversible and is a physical change. This means that physical changes can be identified at macroscopic level. Hence the answer is true.
It’s Benzene, I do believe :) hope this helped, good luck!
