When Zn(OH)2(s) was added to 1.00 L of a basic solution, 1.24×10−2 mol of the solid dissolved. What is the concentration of OH− in the final solution?
To calculate the concentration of OH- in the final solution, we need to first determine the number of moles of OH- that were produced by the dissolution of Zn(OH)2.
The balanced chemical equation for the dissolution of Zn(OH)2 is:
Zn(OH)2 (s) -> Zn2+ (aq) + 2OH- (aq)
From the equation, we can see that for every 1 mole of Zn(OH)2 that dissolves, 2 moles of OH- are produced. Therefore, the number of moles of OH- produced can be calculated using the stoichiometry of the balanced equation.
Given that 1.24×10^-2 mol of Zn(OH)2 was dissolved, we can multiply this by the stoichiometric coefficient of OH- (2) to get the number of moles of OH- produced:
Number of moles of OH- = (1.24×10^-2 mol Zn(OH)2) * (2 mol OH- / 1 mol Zn(OH)2)
= 2.48×10^-2 mol OH-
Next, we need to determine the concentration of OH- in the final solution. Concentration is defined as the number of moles of a solute divided by the volume of the solution in liters:
Concentration (M) = (Number of moles of OH-) / (Volume of solution in liters)
We are given that the volume of the solution is 1.00 L, so we can substitute the values into the formula:
Concentration (M) = (2.48×10^-2 mol OH-) / (1.00 L)
= 2.48×10^-2 M
Therefore, the concentration of OH- in the final solution is 2.48×10^-2 M.