Question

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

Answer 1

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

Explanation:

The given data is as follows.

    Volume of $HClO_{4}$ = 900.0 ml = 0.9 L,

   Molarity of $HClO_{4}$ = 0.18 M,

Hence, we will calculate the number of moles of $HClO_{4}$ 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 $HClO_{4}$ is equal to the number of moles of NaOH.

Also we know that,

         $HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O$

As 1 mole of $HClO_{4}$ 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.