Answer:
Check the explanation
Explanation:
As we know the reaction of EDTA and 
+ and EDTA and 
+
Let us say that the ratio is 1:1
Therefore, the number of moles of + = molarity * volume
= 0.0400M * 0.011L
= 0.00044 moles
Therefore excess EDTA moles = 0.00044 moles
Given , initial moles of EDTA = 0.0430 M * 0.025 L
= 0.001075
Therefore reacting moles of EDTA with 
= 0.001075 - 0.00044 = 0.000675 moles
Let us say that the ratio between and EDTA is 1:1
Therefore moles of = 0.000675 moles
Molarity = moles / volume
= 0.000675 moles / 0.057 L
= 0.011 M (answer).
Answer:
Acetic acid Ka = 1.74 × 10⁻⁵
Trichloroacetic acid Ka = 2 × 10⁻¹
Explanation:
Let's consider the acid dissociation of acetic acid.
CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq)
The pKa of acetic acid is 4.76. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-4.76)
Ka = 1.74 × 10⁻⁵
Let's consider the acid dissociation of trichloroacetic acid.
CCl₃COOH(aq) ⇄ CCl₃COO⁻(aq) + H⁺(aq)
The pKa of trichloroacetic acid is 0.7. The acid dissociation constant (Ka) is:
pKa = -log Ka
- pKa = log Ka
Ka = anti log (-pKa)
Ka = anti log (-0.7)
Ka = 2 × 10⁻¹
B)Buenos Aires was the city near the epicenter :)
Answer:
CuSO4 + 2 NaOH = Cu(OH)2 + Na2SO4
