A 4.00 L solution of KOH is prepared by diuling 400.0 m 0f 0.290 M KOH with warer
[OH-]=
0.290M KOH x (0.400/4) = ?
(OH^-) = (KOH)
To find the concentration of hydroxide ions ([OH-]) in the diluted solution, we can use the dilution formula:
M1V1 = M2V2
Where:
- M1 is the initial concentration of the solution (0.290 M KOH)
- V1 is the initial volume of the solution (400.0 mL or 0.400 L)
- M2 is the final concentration of the solution (which we need to find)
- V2 is the final volume of the solution (4.00 L)
First, we convert the initial volume from milliliters to liters:
V1 = 400.0 mL = 0.400 L
Now we can use the dilution formula to solve for M2:
(0.290 M)(0.400 L) = M2(4.00 L)
M2 = (0.290 M)(0.400 L) / 4.00 L
M2 = 0.029 M
Therefore, the concentration of hydroxide ions ([OH-]) in the diluted solution is 0.029 M.