Calculate [OH−] for 1.90mL of 0.190M NaOH diluted to 1.50L .
(NaOH) = 0.190M x (1.90 mL/1500 mL) = ?
To calculate [OH−] for the given solution, we need to use the concept of dilution. Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent.
First, let's determine the moles of NaOH present in the original solution:
moles of NaOH = volume of solution (in liters) × concentration of NaOH (in moles per liter)
= 0.0019 L × 0.190 mol/L
= 0.000361 moles
Next, let's calculate the new concentration of NaOH after dilution:
new concentration of NaOH = moles of NaOH / volume of dilute solution (in liters)
= 0.000361 moles / 1.5 L
= 0.000241 mol/L
Since NaOH is a strong base, it fully dissociates in water, which means that every mole of NaOH produces one mole of OH− ions. Therefore, the concentration of OH− ions in the diluted solution is also 0.000241 mol/L.
Finally, let's convert the concentration of OH− ions into molarity by dividing by the volume of the dilute solution in liters:
[OH−] = concentration of OH− ions / volume of dilute solution (in liters)
= 0.000241 mol/L / 1.5 L
= 0.00016067 M
Therefore, the concentration of OH− ions in the diluted solution is 0.00016067 M.