Question

What will be the resultant ph of the solution when 200 ml of aqueous solution of hcl with ph 2 is mixed with 300 ml of aqua solution of naoh with ph 12?

Answer 1

Given:

pH of HCl solution = 2

pH of NaOH solution = 12

To find:

pH of new solution = ?

Solution:

pH of acid = Concentration of $H^+$ ions

pH of base = Concentration of $OH^-$ ions

Accordingly,

pH 2.0 of HCl corresponds to hydrogen ion concentration of 0.01 M.

∴Number of moles of hydrogen ions present = 0.01× 0.2 = 0.0020

pH 12 of NaOH correspond to hydroxide ion concentration of 0.01 M

∴Number of moles of hydroxide ions present = 0.01 × 0.3 = 0.0030

As acid and base are reacting, neutralizing occurs. Therefore out of 0.0030 moles 0.0020 moles are used.

Thus only 0.0010 moles are left in the solution.

∴ Concentration of hydroxide ion = $\frac{0.0010}{0.5} = 0.0020M$

pOH = -log[$OH^-$]

pOH = -log (0.0020)

pOH = 2.699

pH of the solution = 14-pOH

pH = 14-2.699

pH = 11

Conclusion:

The pH of the solution newly obtained is 11

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