HBr and HF are both monoprotic Arrhenius acids—that is, in aqueous solution, they dissociate and ionize to give hydrogen ions. A strong acid ionizes completely; a weak acid ionizes partially.
In this case, HBr, being a strong acid, would ionize completely in water to yield H+ and Br- ions. However, HF, being a weak acid, would ionize only to a limited extent: some of the HF molecules will ionize into H+ and F- ions, but most of the HF will remain undissociated.
pH is, by definition, a measurement of the concentration of hydrogen ions in solution (pH = -log[H+]). A higher concentration of hydrogen ions gives a lower pH, while a lower concentration of hydrogen ions gives a higher pH. At 25 °C, a pH of 7 indicates a neutral solution; a pH less than 7 indicates an acidic solution; and a pH greater than 7 indicates a basic solution.
If we have equal concentrations of HBr and HF, then the HBr solution will have a greater concentration of hydrogen ions in solution than the HF solution. Consequently, the pH of the HBr solution will be less than the pH of the HF solution.
Choice A is incorrect: Strong acids like HBr dissociate completely, not partially.
Choice B is incorrect: While the initial concentration of HBr and HF are the same, the H+ concentration in the HBr solution is greater. Since pH is a function of H+ concentration, the pH of the two solutions cannot be the same.
Choice C is correct: A greater H+ concentration gives a lower pH value. The HBr solution has the greater H+ concentration. Thus, the pH of the HBr solution would be less than that of the HF solution.
Choice D is incorrect for the reason why choice C is correct.
Explanation:
will dissociate into ions as follows.
Hence, 
for this reaction will be as follows.
![K_{sp} = [Pb^{2+}][Br^{-}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BPb%5E%7B2%2B%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
We take x as the molar solubility of when we dissolve x moles of solution per liter.
Hence, ionic molarities in the saturated solution will be as follows.
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= ![[Pb^{2+}]_{o}](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D)
+ x
![[Br^{-}]^{2}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
= ![[Br^{-}]_{o}](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D)
+ 2x
So, equilibrium solubility expression will be as follows.
= ![([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}](https://tex.z-dn.net/?f=%28%5BPb%5E%7B2%2B%7D%5D_%7Bo%7D%20%2B%20x%29%28%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x%29%5E%7B2%7D)
Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains = 0.10 and there will be no lead ions. So, = 0
So, ![[Br^{-}]_{o} + 2x](https://tex.z-dn.net/?f=%5BBr%5E%7B-%7D%5D_%7Bo%7D%20%2B%202x)
will approximately equals to .
Hence, ![K_{sp} = x[Br^{-}]^{2}_{o}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20x%5BBr%5E%7B-%7D%5D%5E%7B2%7D_%7Bo%7D)
x = 
M
Thus, we can conclude that molar solubility of is M.
The answer is B as isotopes are different versions of the same chemical element containing the same amount of protons and electrons but different amounts of neutrons.
Answer:
There are 5 significant digits in 0.23100.
Explanation:
This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.
