Answer:
Molarity of Sr(OH)₂ = 0.47 M
Explanation:
Given data:
Volume of Sr(OH)₂ = 15.0 mL
Volume of HCl = 38.5 mL (0.0385 L)
Molarity of HCl = 0.350 M
Concentration/Molarity of Sr(OH)₂ = ?
Solution:
Chemical equation:
Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O
Number of moles of HCl:
Molarity = number of moles/ volume in L
0.350 M = number of moles/0.0385 L
Number of moles = 0.350 mol/L× 0.0385 L
Number of moles = 0.0135 mol
Now we will compare the moles of HCl with Sr(OH)₂.
HCl : Sr(OH)₂
2 : 1
0.0135 : 1/2×0.0135 = 0.007 mol
Molarity/concentration of Sr(OH)₂:
Molarity = number of moles / volume in L
Molarity = 0.007 mol /0.015 L
Molarity = 0.47 M
The required molarity of 15.0 mL [tex]Sr(OH)_2[/tex] solution is 0.449 M and the further calculation can be defined as follows:
Given that:
Volume of [tex]Sr(OH)_2[/tex] solution = 15.0 mL
n-factor of [tex]Sr(OH)_2[/tex] solution = 2
Molarity of HCl solution = 0.350 M
Volume of HCl solution = 38.5 mL
n-factor of HCl solution = 1
To find:
Molarity of [tex]Sr(OH)_2[/tex] solution required.
Formula used to calculate the molarity of [tex]Sr(OH)_2[/tex] solution is:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where, [tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acidic solution while [tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of basic solution.
Put the values in the above formula:
[tex]1\times 0.350\times 38.5=2\times M_2\times 15.0\\\\M_2=\frac{1\times 0.350\times 38.5}{2\times 15.0}\\\\M_2=0.449M[/tex]
So, the required molarity of [tex]Sr(OH)}2[/tex] solution is 0.449 M.
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