calculate the volume of 0.50 M NaOH needed to saturate 100.0 ml of 0.50 M Zn(NO3)2 with Zn(OH)2
To calculate the volume of 0.50 M NaOH needed to saturate 100.0 ml of 0.50 M Zn(NO3)2 with Zn(OH)2, you need to determine the stoichiometry of the reaction between Zn(NO3)2 and NaOH.
The balanced equation for the reaction between Zn(NO3)2 and NaOH is as follows:
Zn(NO3)2 + 2NaOH -> Zn(OH)2 + 2NaNO3
From the balanced equation, you can see that 1 mole of Zn(NO3)2 reacts with 2 moles of NaOH to produce 1 mole of Zn(OH)2.
First, calculate the number of moles of Zn(NO3)2 in 100.0 ml of a 0.50 M solution. To do this, you can use the equation:
moles = concentration (M) x volume (L)
Given that the volume is 100.0 ml (0.100 L) and the concentration is 0.50 M, you can calculate the moles of Zn(NO3)2:
moles of Zn(NO3)2 = 0.50 M x 0.100 L = 0.050 moles
Since 1 mole of Zn(NO3)2 reacts with 2 moles of NaOH, the number of moles of NaOH required to saturate the Zn(NO3)2 solution is also 0.050 moles.
Finally, you can calculate the volume of 0.50 M NaOH needed to supply 0.050 moles using the equation:
volume (L) = moles / concentration (M)
Given the concentration of NaOH is 0.50 M, you can calculate the volume:
volume of NaOH = 0.050 moles / 0.50 M = 0.10 L or 100 ml
Therefore, you would need 100 ml (0.10 L) of 0.50 M NaOH to saturate 100.0 ml of 0.50 M Zn(NO3)2 with Zn(OH)2.