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24. 00 ml of a 0. 25 m naoh solution is titrated with 0. 10m hcl. What is the ph of the solution after 24. 00 ml of the hcl has been added?.

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The pH of the solution is 12.88.

Calculation:

millimoles of NaOH = volume x Molarity

                                 = 24.00 mL x 0.25 M

                                 = 6.00 mmole

millimoles of HCl = 24.00 mL x 0.10 M

                            = 2.40 mmole

Total volume = 48.00 mL

                           

                            NaOH + HCl   →     NaCl + H2O

initial                       6.00      0               0         0

added                                 2.40

change                   -2.40  -2.40      +2.40  +2.40

equilibrium               3.60    0           2.40    2.40

The pH is determined by the excess of NaOH present.

[NaOH] = millimoles/volume

            = 3.60/48.00

            = 0.075 M

pOH = -log (OH⁻).

        = -log (0.075)

        = 1.12

Then

pH + pOH = pKw = 14.00.

pH = 14 - pOH

pH = 14 - 1.12

pH = 12.88

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