Question

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?\

Answer 1

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

Answer 2

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