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
Sorry, haven't came across that as of yet.
Answer:
6 days
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Half life (t½) =?
Next, we shall determine the decay constant. This can be obtained as follow:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Decay constant (K) =?
Log (N₀/N) = kt / 2.303
Log (100/6.25) = k × 24 / 2.303
Log 16 = k × 24 / 2.303
1.2041 = k × 24 / 2.303
Cross multiply
k × 24 = 1.2041 × 2.303
Divide both side by 24
K = (1.2041 × 2.303) / 24
K = 0.1155 /day
Finally, we shall determine the half-life of the isotope as follow:
Decay constant (K) = 0.1155 /day
Half life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 0.1155
t½ = 6 days
Therefore, the half-life of the isotope is 6 days
