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.
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
C is the answer.
The temperature T<span> in degrees Celsius (°C) is equal to the temperature </span>T<span> in Kelvin (K) minus 273</span>°.
Answer:
Cold
Explanation:
Because 0°C And below or 32°F Refers To Cold Weather
