It's a hazard symbol for irritant.
It could be a tide pool but this is pretty vague
Answer:
- <u>Yes,</u> <em>all titrations of a strong base with a strong acid have the same pH at the equivalence point.</em>
This <u>pH is 7.</u>
Explanation:
<em>Strong acids</em> and <em>strong bases</em> ionize completely in aqueous solutions. The ionization of strong acids produce hydronium ions, H₃O⁺, and the ionization of strong bases produce hydroxide ions, OH⁻.
Since the ionization of strong acids and bases progress until completion, there is not reverse reaction.
The definition of pH is pH = - log [H₃O⁺]. Acids have low pH (below 7, and greater than 0) and bases have high pH (above 7 and less than 14). Neutral solutions have pH = 7.
Acid-base titrations are a method to determine the concentration of an acid from the known concentration of a base, or the concentraion of a base from the known concentration of an acid.
The<em> equivalence point</em> of the titration is the point at which the the number of moles of hydronium ions and hydroxide ions are equal.
Then, at that point, the hydronium and hydroxide ions will be in the stoichiometric proportion to form a neutral solution, i.e. the pH of the solution wiill be 7.
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
