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Answer:
1. 0.02 M
2. 0.01 M
3. 4×10⁻⁶
Explanation:
We know that V₁S₁ = V₂S₂
1.
Concentration of HCl = 0.05 M
end point comes at = 10 ml
So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M
2.
2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)
[Ca²⁺] = 0.02 ÷ 2 = 0.01 M
3.
= [Ca²⁺(aq)] [OH⁻(aq)]²
Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)
= [0.01 × (0.02)²] = 4×10⁻⁶
4.
If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.
5.
value describes the solubility of a particular ionic compound. The higher the value, the higher the Solubility will be.
6.
This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion
1) Content of Ca (2+) ions
Calcium chloride = CaCl2
Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)
=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)
Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution
M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2
=> 0.0825 mol Ca(2+)
2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)
formula of phospahte ion: PO4 (3-)
molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2
Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)
=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)
3) Content of Mg(2+) ions
Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)
Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)
number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution
n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2
ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)
4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)
2PO4(3-) + 3Mg(2+) = Mg3(PO4)2
=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)
=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)
5) Total number of moles of PO4(3-)
0.055 mol + 0.16 mol = 0.215 mol
6) Sodium phosphate
Sodium phosphate = Na3(PO4)
Na3PO4 ---> 3Na(+) + PO4(3-)
=> 1 mol Na3PO4 : 1 mol PO4(3-)
=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4
mass in grams = number of moles * molar mass
molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol
=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g
Answer: 35.26 g of sodium phosphate
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