Question

20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is \(\_\_\_\_\_\) x \(10^{-2}\)M. (Nearest integer) 

Hint:

To calculate the final concentration when mixing two solutions, use the dilution equation and ensure you add up the volumes and concentrations properly.

Answer 1

Correct Answer: 5.7

We can calculate the final concentration of the NaOH solution using the dilution formula: \[ C_1 V_1 + C_2 V_2 = C_f V_f \] 
Where: - \( C_1 = 2 \, \text{M} \) (concentration of first solution), - \( V_1 = 20 \, \text{mL} \) (volume of first solution), - \( C_2 = 0.5 \, \text{M} \) (concentration of second solution), - \( V_2 = 400 \, \text{mL} \) (volume of second solution), - \( C_f \) is the final concentration, and - \( V_f = V_1 + V_2 = 20 + 400 = 420 \, \text{mL} \). 

Now, substitute the values: \[ (2 \times 20) + (0.5 \times 400) = C_f \times 420 \] \[ 40 + 200 = C_f \times 420 \] \[ C_f = \frac{240}{420} = 0.571 \, \text{M} \] 

Thus, the final concentration is approximately \( 0.57 \, \text{M} \), or \( 5.7 \times 10^{-2} \, \text{M} \).