You are given the neutralization of acetic acid with sodium hydroxide. Also, you are given the k for acetic acid, which is 1.8 x 10⁻⁵. You are asked to find the<span> approximate value of the equilibrium constant, kn, for the neutralization. We will have a reaction of both acetic acid and sodium hydroxide.
CH</span>₃COOH + NaOH → CH₃COONa + H₂O
which comes from
CH₃COOH → CH₃COO⁻ + H⁺
H⁺ + OH⁻ → H₂O<span>
</span>The k for water is always 1.0 x 10¹⁴. The Ksp for the reaction will be
<span>
Ksp = [</span>CH₃COOH][H₂O]
Ksp = (1.8 x 10⁻⁵)(1.0 x 10¹⁴)
<span>Ksp = 1.8 x 10</span>⁹
Acceleration (change of movement) of an object is dependent on the net force acting on it and its mass.
The relationship between acceleration (a), force (f) and mass (m) is given by the Second Lay of Newton: f = m * a.
Note that the force, f, is the net force acting on the objetct.
I hope this help you.
Answer:
pH = 4.57
Explanation:
pH = pKa + log ([OAc⁺]/[HOAc])
Ka(HOAc) 1.8 x 10⁻⁵ => pKa = -log(1.8 x 10⁻⁵) = 4.74
[OAc⁻] = 0.20M
[HOAc] = 0.30M
pH = 4.74 + log([0.20]/[0.30]) = 4.47 + (-0.17) = 4.57
1. 3 significant figures
2. 5 significant figures
3. 3 significant figures
4. 4 significant figures
5. 5 significant figures
6. 2 significant figures
7. 4 significant figures
8. 1 significant figure
9. 4 significant figures
