Assume that 268mL of 5.00M NaOH are diluted to a final volume of 574mL with water. What is the final molarity of NaOH (FM=40.0).
mL1 x M1 = mL2 x M2
258 x 5.00 = 574 x M2.
To find the final molarity of NaOH, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial molarity of NaOH
V1 = initial volume of NaOH
M2 = final molarity of NaOH (what we are trying to find)
V2 = final volume of NaOH (obtained by dilution)
Given:
M1 = 5.00M
V1 = 268mL
V2 = 574mL
Plugging in these values into the formula, we get:
(5.00M)(268mL) = M2(574mL)
Now, we can solve for M2:
M2 = (5.00M)(268mL) / (574mL)
Calculating this, we get:
M2 ≈ 2.33M
So, the final molarity of NaOH after dilution is approximately 2.33M.