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 · 2019-10-15T02:13:48-04:00

Answer: 0.0218 M

Explanation:

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

\(Molarity=\frac{n}{V_s}\)

where,

n= moles of solute

\(V_s\) = volume of solution in Liters

\(Zn(OH)_2\rightarrow Zn^{2+}+2OH^-\)

According to stoichiometry:

1 mole of \(Zn(OH)_2\) gives = 2 moles of \(OH^-\)

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

\(Molarity=\frac{2.18\times 10^{-2}}{1.00L}=0.0218M\)

Thus concentration of \(OH^-\) in the final solution will be 0.0218 M