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 _______ ×10⁻² M. (Nearest integer)

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Moles of NaOH $=0.020	imes2+0.400	imes0.5=0.04+0.20=0.24$ mol in $0.420$ L. $M=\frac{0.24}{0.42}=0.571=57.1	imes10^{-2}$ M, i.e. 57. Answer: 57.

Q. 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 _______ ×10⁻² M. (Nearest integer)

Explanation

Reviewed by Dr. Abubacker

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