M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
We know that:
M1 = 0.211 M
V1 = 55 mL
V2 = 125 mL
Substituting these values into the formula, we get:
0.211 M * 55 mL = M2 * 125 mL
Solving for M2, we get:
M2 = (0.211 M * 55 mL) / 125 mL
M2 = 0.089 M
Therefore, the concentration of NaOH in the diluted solution is 0.089 M.
