Answer:
The pH of a solution is simply a measure of the concentration of hydrogen ions,
H
+
, which you'll often see referred to as hydronium cations,
H
3
O
+
.
More specifically, the pH of the solution is calculated using the negative log base
10
of the concentration of the hydronium cations.
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
−
log
(
[
H
3
O
+
]
)
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Now, we use the negative log base
10
because the concentration of hydronium cations is usually significantly smaller than
1
.
As you know, every increase in the value of a log function corresponds to one order of magnitude.
Explanation:
Strontium (Sr). Elements in the same group of the periodic table have similar characteristics.
C=0.10 mol/l
pH=-lg[H⁺]
HCl = H⁺ + Cl⁻
pH=-lgc
pH=-lg0.10=1.0
pH=1.0
The amount
per 100 g is:
38.7 %
calcium = 38.7g Ca / 100g compound = 38.7g
19.9 %
phosphorus = 19.9g P / 100g compound = 19.9g
41.2 %
oxygen = 41.2g O / 100g compound = 41.2g
The molar amounts of calcium,
phosphorus and oxygen in 100g sample are calculated by dividing each element’s
mass by its molar mass:
Ca = 38.7/40.078
= 0.96
P = 19.9/30.97
= 0.64
O = 41.2/15.99
= 2.57
C0efficients
for the tentative empirical formula are derived by dividing each molar amount
by the lesser value that is 0.64 and in this case, after that multiply wih 2.
Ca = 0.96 /
0.64 = 1.5=1.5 x 2 = 3
P = 0.64 /
0.64 = 1 = 1x2= 2
O = 2.57 /
0.64 = 4= 4x2= 8
Since, the
resulting ratio is calcium 3, phosphorus 2 and oxygen 8
<span>So, the
empirical formula of the compound is Ca</span>₃(PO₄)₂
1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. Na₂CO₃ as a limiting reactant
<h3>Further explanation</h3>
Given
Reaction
2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃
Required
mol ratio
Limiting reactant
Solution
The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)
1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3
2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67
Na₂CO₃ as a limiting reactant (smaller)
