To calculate the pH of this solution, we use the
Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where,
[A-] = Molarity of the conjugate base =
CH3COO- = 0.29 M<span>
<span>[HA] = Molarity of the weak acid = CH3COOH = 0.18 M</span></span>
pKa = dissociation constant of the weak acid =
4.75
When KOH is added to the buffer, the chemical
reaction is:
CH3COOH + KOH = CH3COO-K+ + H2O
Therefore when 0.0090 mol KOH is added, 0.0090
mol acid is neutralized, and 0.0090 mol CH3COO- is produced.
[CH3COO-] = [0.0090 mol + 0.375 L (0.29 mol/L) ]
/ 0.375 L = 0.314 M
[CH3COOH] = [-0.0090 mol + 0.375 L (0.18 mol/L) ]
/ 0.375 L = 0.156 M
Going back to Henderson-Hasselbalch
equation:
pH = 4.75 + log (0.314 / 0.156)
<span>pH = 5.054</span>
The correct answer should be letter choice C) There is an equal number of each type of atom on the reactant and product side. Its C) because if its trying to conserve mass, than that means save or be equally balanced so it should be even for both sides.
A reducing agent is one which is oxidised in the reaction itself. When you take into account the oxidation numbers you will see that the Cl- ions are oxidised from an oxidation number of -1 to 0 in Cl2. Therefore Cl- ions are the reducing agent.
