Data:
log10(1.8) ≈ 0.25
Solving:
[tex]pH = - log[H_{3} O^+][/tex]
[tex]pH = - log1.840 (mol/L)[/tex]
[tex]pH = - log1.8*10^3[/tex]
[tex]pH = 3 - 0.25[/tex]
[tex]\boxed{\boxed{pH = 2.75}}\end{array}}\qquad\quad\checkmark[/tex]
Note:. The pH <7, then we have an acidic solution.
