p
K
a
=
5.4
pH
=
3.7
Explanation:
p
K
a
is simply
−
log
10
(
K
a
)
. It is a constant at any temperature and does not depend on the molarity of the solution.
p
K
a
=
−
log
10
(
K
a
)
=
−
log
10
(
4.0
×
10
−
6
)
=
5.4Explanation:
gills are most likely found in all fish
Answer is: 4) The same subscripts are on each side of the equation.
For example, balanced chemical reaction:
2Mg + O₂ → 2MgO.
1) The same number of atoms is on each side of the equation: two magnesium atoms and two oxgen atoms.
2) The formulas for all substances are correct: in magnesium oxide (MgO), magnesium has oxidation number +2 and oxygen -2, so formula is good, because compound must be neutral.
3) The same mass is represented on each side of the equation: because there is same number of atoms, the mass is the same.
4) The same subscripts are on each side of the equation: oxygen does not have same subscripts.
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
Answer:
Lithium has a melting point of 180.54 C, a boiling point of 1342 C, a specific gravity of 0.534 (20 C), and a valence of 1. It is the lightest of the metals, with a density approximately half that of water. Under ordinary conditions, lithium is the least dense of the solid elements.
Appearance: soft, silvery-white metal
Atomic Number: 3
Atomic Radius (pm): 155
