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.
Ionic bond is a chemical bond formed by the complete transfer of electrons between two atoms. The atom that loses electrons gains a positive charge (cation) and that which accepts electrons gains a negative charge (anion). Now, electronegativity is a parameter that measures the tendency of an atom to accept electrons. In the context of ionic bonding, two elements which show a significant difference in their electronegativity values form ionic bonds.
In the given examples, the difference in electronegativity is greatest between K and Br i.e. 0.8 and 2.8 respectively with a difference of 2.0. This also makes sense since K and Br are on the extreme ends of the periodic table. Hence, potassium with a valence electron configuration of 4s1 will lose its s electron to Br (4s24p6) and form an ionic molecule K⁺Br⁻
Ans E) potassium and bromine
The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced
It is Vitamin D, hope that helps
