Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Answer:
<em>Valency is the combining power of an element. Elements in the same group of the periodic table have the same valency. The valency of an element is related to how many electrons are in the outer shell. The noble gases have the valency 0 as they do not usually combine with other elements.</em>
Explanation:
HOPE IT HELPS
Answer:
0.5 M
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 80 g
Volume of solution = 4 L
Molarity =?
Next, we shall determine the number of mole in 80 g of NaOH. This can be obtained as follow:
Mass of NaOH = 80 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 80 / 40
Mole of NaOH = 2 moles
Finally, we shall determine the molarity of the solution. This can be obtained as follow:
Mole of NaOH = 2 moles
Volume of solution = 4 L
Molarity =?
Molarity = mole / Volume
Molarity = 2/4
Molarity = 0.5 M
Therefore, the molarity of the solution is 0.5 M.
Answer:
d. K<1 E∘cell is negative
Explanation:
Since E⁰ = negative , ΔG = -nFE⁰ = -nF -ve = +ve.
Also, ΔG = -RTlnK
K = exp(-RTΔG)
Since ΔG = +ve, -RTΔG = -ve
K = 1/exp(RTΔG) < 1.
So our answer is E⁰ cell is negative and K < 1
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
