What is the pH of the solution obtained by mixing 30 mL of .25 M HCL in 30 mL of .125 M NaOH

The pH of the resulting solution is 1.2

HCl(aq) + NaOH(aq) > NaCl(aq) + H2O(l)

We are mixing a strong acid and a strong base. The mole ratio as we can see in this reaction is 1:1.

Let us now obtain the number of moles of each reactant;

n(HCl) = 30/1000 × 0.25 = 0.0075 moles

n(NaOH) = 30/1000  × 0.125 = 0.00375 moles

We can see that there are twice as much moles of HCl than moles of NaOH. Hence, NaOH is the limiting reactant.

Hence, 0.00375 moles moles of NaOH reacts with 0.00375 moles of HCl and 0.00375 moles of HCl is left over.

Total volume of solution = 30 mL + 30 mL= 60 mL or 0.06 L

Concentration of H3O^+ left in solution = 0.00375 moles/0.06 L

= 0.0625 M

pH = -log[ H3O^+]

pH = -log[0.0625 M]

pH = 1.2

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pH after mixing = 1.2

Explanation:

HCl + NaOH => NaCl + H₂O

Will show only the reactants for the solution set ...

Given 30ml(0.25M HCl) + 30ml(0.125M NaOH)

=> 0.030(0.25)mole HCl + 0.030(0.125)mole NaOH

=> 0.0075mole HCl + 0.00345mol NaOH

moles HCl neutralized = moles NaOH added into solution (1:1 reaction ratio)

=> (0.0075 - 0.00375)mol HCl remains in excess in 60ml of solution

=> (0.00375mol/0.060L)HCl = 0.0625M in HCl

pH = -log[HCl] = -log[H⁺] = -log(0.0625) = -(-1.2) = 1.2