Part 1: Calculate the percent ionization of 0.0075 m butanoic acid (Ka=1.5x10^-5)
C4H8O2(aq) +H2O(l) → C4H7O2(aq) + H3O +
initial 0.0075 0 0
change -X +X +X
final 0.0075-X X X
when Ka is relative smaller to the intial concentration of the acid so we can assume that 0.0075-X≈ 0.0075 by substitution in Ka formula:
Ka = [C4H7O2][H3O+] / [C4H8O2]
1.5x10^-5 = X*X / 0.0075
X^2 = 1.125x10^-7
X= 0.00034 m =3.4 x 10 ^-4 m
∴ [C4H7O2] = [H3O+] = 3.4X10^-4
∴ percent ionization = [H+ equlibrium]/[acid initial] *100
= 3.4X10^-4/0.0075 * 100
= 4.5 %
part 2) calculate the percent ionization of 0.0075m butanoic acid in a solution containing 0.085m sodium butanoic?
C4H8O2(aq) + H2O(l) ↔ C4H7O2(aq) + H3O+
initial 0.0075 0 0.085
change -X +X +X
final 0.0075-X X 0.085+X
we can assume that 0.0075-X≈ 0.0075 & 0.085+X ≈ 0.085
∴Ka = (X*(0.085)) / (0.0075)
(1.5x10^-5)*0.0075 = 0.085X
∴X = 1.3x 10^-6
∴ percent ionization = (1.3x10^-6)/0.0075 * 100 = 0.017 %
To become positively charged it loses electrons.
To become negatively charged it gains electrons.
Electrons have a negative charge.
Hope that helps!
Explanation:
Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured vapour. Its properties are intermediate between those of chlorine and iodine. Isolated independently by two chemists, Carl Jacob Löwig (in 1825) and Antoine Jérôme Balard (in 1826), its name was derived from the Ancient Greek βρῶμος ("stench"), referring to its sharp and disagreeable smell.
Bromine, 35Br
In a controlled experiment the scientist controls only one side of the variables.