When zn(oh)2(s) was added to 1.00 l of a basic solution, 1.09×10−2 mol of the solid dissolved. what is the concentration of oh− in the final solution?\

The concentration of the basic solution is determined by:

N = (number of moles / volume of solution)

number of moles = 1.09 x 10^-2 mol
volume of solution = 1 liter
N of basic solution = 1.09 x 10^-2 mol / 1 liter
N = 1.09 x 10^-2 mol/L

The initial concentration of Zn (OH)2 is 0; the basic solution is 1.09x10^-2 M, then the concentration of OH in the final solution is 1.09x10^-2 M

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kobenhavn kobenhavn · Answer 2 · 2019-10-15T08:13:48+02:00

Answer: 0.0218 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]Molarity=\frac{n}{V_s}[/tex]

where,

n= moles of solute

[tex]V_s[/tex] = volume of solution in Liters

[tex]Zn(OH)_2\rightarrow Zn^{2+}+2OH^-[/tex]

According to stoichiometry:

1 mole of [tex]Zn(OH)_2[/tex] gives = 2 moles of [tex]OH^-[/tex]

Thus [tex]1.09\times 10^{-2}[/tex] gives = [tex]\frac{2}{1}\times 1.09\times 10^{-2}=2.18\times 10^{-2}moles of [tex]OH^-[/tex]

[tex]Molarity=\frac{2.18\times 10^{-2}}{1.00L}=0.0218M[/tex]

Thus concentration of [tex]OH^-[/tex] in the final solution will be 0.0218 M