What is the pH of the solution formed by mixing 40 mL of 0.1 M HCl with 10 mL of 0.45 N NaOH?

Question

At \(25^\circ \text{C}\), the ionic product of water is \(10^{-14}\). The free energy change (\(\Delta G^\circ\)) for the self-ionization of water in kcal mol\(^{-1}\) is:

Answer 1

meq acid=0.1x40=4. meq base=0.45x10=4.5. Excess base=0.5 meq/50mL=0.01N=[OH-]. pOH=2, pH=12.

\[\boxed{pH = 12}\]

What is the pH of the solution formed by mixing 40 mL of 0.1 M HCl with 10 mL of 0.45 N NaOH?

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The Correct Option is D

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