Answer: HCI + KOH → KCI + H20
Explanation:
HCI(aq) + KOH(aq) → KCI(aq) + H20(l)
Acid + base → Salt + Water.
The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.
This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.
We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
pH + pOH = 14
therefore pH = 14 - 4.42
pH = 9.58
6.02 times 10 to the 23 power, which is Avogadros number
The reaction must be a + b --> c
Then you can predict a reaction rate, r o the type r = k * a^n * b^m
Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.
Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.
Answer: r = k*a
Answer: mL of water that should be add
3.826mL
Explanation:
Check attachment
