Let's investigate the substances involved in the reaction first. The compound <span>CH3NH3+Cl- is a salt from the weak base CH3NH2 and the strong acid HCl. When this salt is hydrated with water, it will dissociate into CH3NH2Cl and H3O+:
CH3NH3+Cl- + H2O </span>⇒ CH3NH2Cl + H3O+
Nest, let's apply the ICE(Initial-Change-Equilibrium) table where x is denoted as the number of moles used up in the reaction:
CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+
Initial 0.51 0 0
Change -x +x +x
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Equilibrium 0.51 - x x x
Then, let's find the equilibrium constant of the reaction. Since the reaction is hydrolysis we use KH, which is the ratio of Kw to Ka or Kb. Kw is the equilibrium constant for water hydrolysis which is equal to 1×10⁻¹⁴. Since the salt comes from the weak base, we use Kb. Since pKb = 3.44, then. 3.44 = -log(Kb). Thus, Kb = 3.6307×10⁻⁴
KH = Kw/Kb = (x)(x)/(0.51 - x)
1×10⁻¹⁴/ 3.6307×10⁻⁴ = x²/(0.51-x)
x = 3.748×10⁻⁶
Since x from the ICE table is equal to the equilibrium concentration of H+, we can find the pH of the aqueous solution:
pH = -log(H+) = -log(x)
pH = -log ( 3.748×10⁻⁶)
pH = 5.43
Answer:
Water has a pH of 7 - its not acid or base (its neutral {in the middle})
Explanation:
Since the pH scale maximum number is 14, the middle number - is 7 - which is where water is
Answer:
1.5 Molar solution in KCl
Explanation:
Definition of molarity => moles of solute / Volume of solution in Liters
Given 3.00 moles of solute dissolved in 2Liters of solution
Molarity of [KCl] = 3.00 moles KCl / 2Liters solution = 1.5 Molar solution in KCl
Answer:
When the solution (with phenolphthalein) changes to colorless
Explanation:
When titrating with HCl is common to add phenolphthalein as an acid-base indicator.
Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.
So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.
