Respuesta :
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 [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,
Molarity of [tex]HClO_{4}[/tex] = 0.18 M,
Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] 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 [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.
Also we know that,
[tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]
As 1 mole of [tex]HClO_{4}[/tex] 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.
