Answer: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.
The concentration of OH- = 0.473 M.
Balanced chemical equation:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
[tex]C_A*V_A/C_B*V_B = N_A/N_B[/tex]
Where,
[tex]N_A[/tex] is the number of mole of acid = 2
[tex]N_B[/tex] is the number of mole of base = 1
[tex]C_A[/tex] is the concentration of acid = 1M
[tex]C_B[/tex] is the concentration of base = to be calculated
[tex]V_A[/tex] is the volume of acid = 23.65 ml
[tex]V_B[/tex] is the volume of base = 25mL
On substituting the values:
[tex]1*23.65/C_B*25 = 2/1\\\\C_B =1*23.65*1/25*2\\\\C_B = 0.473 M[/tex]
Thus, the concentration of OH- = 0.473 M.
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