Answer:
1. 0.00352 M
2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3. 0.00534 M
Explanation:
1.
Mass of strontium hydroxide= 10.45 g
Volume of solution = 41.00 ml
Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles
Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M
2.
2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3.
Concentration of acid CA= the unknown
Volume of acid VA= 31.5 ml
Concentration of base CB= 0.00352 M
Volume of base VB= 23.9 ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
CA= 0.00352 × 23.9 ×2/31.5 ×1
CA= 0.00534 M
A. The molarity of the Sr(OH)₂ solution is 2.09 M
B. The balanced equation for the reaction is
2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O
C. The molarity of the acid, HNO₃ is 3.17 M
A. Determination of the molarity of the Sr(OH)₂ solution
Mass of Sr(OH)₂ = 10.45 g
Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol
Mole = mass / molar mass
Mole of Sr(OH)₂ = 10.45 / 122
Mole of Sr(OH)₂ = 0.0857 mole
Volume = 41 mL = 41 / 1000 = 0.041 L
Molarity = mole / Volume
Molarity of Sr(OH)₂ = 0.0857 / 0.041
B. The balanced equation for the reaction.
C. Determination of the molarity of the acid, HNO₃.
From the balanced equation above,
The mole ratio of the acid, HNO₃ (nA) = 2
The mole ratio of the base, Sr(OH)₂ (nB) = 1
From the question given above,
Volume of base, Sr(OH)₂ (Vb) = 23.9 mL
Molarity of base, Sr(OH)₂ (Mb) = 2.09 M
Volume of acid, HNO₃ (Va) = 31.5 mL
MaVa / MbVb = nA/nB
(Ma × 31.5) / (2.09 × 23.9) = 2
(Ma × 31.5) / 49.951 = 2
Cross multiply
Ma × 31.5 = 49.951 × 2
Ma × 31.5 = 99.902
Divide both side by 31.5
Ma = 99.902 / 31.5
Thus, molarity of the acid, HNO₃ is 3.17 M
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