A 900.0 mLmL sample of 0.18 MHClO4MHClO4 is titrated with 0.27 MLiOHMLiOH. Determine the pHpH of the solution after the addition of 600.0 mLmL of LiOHLiOH (this is the equivalence point). A 900.0 sample of 0.18 is titrated with 0.27 . Determine the of the solution after the addition of 600.0 of (this is the equivalence point). 11.24 13.03 2.76 0.97 7.00

pH of the solution after the addition of 600.0 ml of LiOH is 7.

Explanation:

The given data is as follows.

    Volume of = 900.0 ml = 0.9 L,

   Molarity of = 0.18 M,

Hence, we will calculate the number of moles of as follows.

       No. of moles = Molarity × Volume

                             = 0.18 M × 0.9 L

                             = 0.162 moles

Volume of NaOH = 600.0 ml = 0.6 L

Molarity of NaOH = 0.27 M

No. of moles of NaOH = Molarity × Volume

                                     = 0.27 M × 0.6 L

                                     = 0.162 moles

This shows that the number of moles of is equal to the number of moles of NaOH.

Also we know that,

As 1 mole of reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.

This means that pH = 7.

Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.

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