10 mL of 2(M) NaOH solution is added to 200 mL of 0.5(M) of NaOH solution. What is the final concentration?

To find the final concentration after mixing 10 mL of 2 M NaOH solution with 200 mL of 0.5 M NaOH solution, we need to calculate the total moles of NaOH and then divide by the total volume of the mixture.

First, calculate the moles of NaOH from the first solution. Molarity is moles per liter, so convert the volume from milliliters to liters. The volume of the first solution is 10 mL, which is $$10 \div 1000 = 0.01$$ liters. The molarity is 2 M, so moles from first solution = molarity × volume = $$2 \times 0.01 = 0.02$$ moles.

Next, calculate the moles of NaOH from the second solution. The volume is 200 mL, which is $$200 \div 1000 = 0.2$$ liters. The molarity is 0.5 M, so moles from second solution = $$0.5 \times 0.2 = 0.1$$ moles.

Now, add the moles from both solutions to get the total moles of NaOH. Total moles = moles from first solution + moles from second solution = $$0.02 + 0.1 = 0.12$$ moles.

The total volume of the mixture is the sum of the volumes of both solutions. Total volume = $$10 \text{ mL} + 200 \text{ mL} = 210$$ mL. Convert this to liters: $$210 \div 1000 = 0.21$$ liters.

The final concentration is the total moles divided by the total volume in liters. Final concentration = $$\frac{0.12}{0.21}$$ M.

Simplify this fraction: $$\frac{0.12}{0.21} = \frac{120}{210}$$ (multiplying numerator and denominator by 1000 to eliminate decimals). Reduce the fraction by dividing both numerator and denominator by 30: $$\frac{120 \div 30}{210 \div 30} = \frac{4}{7} \approx 0.5714$$ M. Rounding to two decimal places gives 0.57 M.

Comparing with the options, 0.57 M corresponds to option A.

Hence, the correct answer is Option A.