Answer:
[H₃O⁺] = 6.31 x 10⁻² mol·L⁻¹
Explanation:
pH = -log[H₃O⁺], so [tex][\text{H}_{3}\text{O}^{+}] = 10^{- \text{pH}} \text{ mol/L}[/tex]
[tex][\text{H}_{3}\text{O}^{+}] = 10^{-1.20}\text{ mol/L} = 6.31 \times 10^{-2}\text{ mol/L}[/tex]
Answer:
[tex]6.31*10^{-2}[/tex]
Explanation:
Hello! To calculate the pH we have the following formula:
ph = -log [H3O +]
In this case we have the pH value so the formula is
[H3O +] = [tex]10^{-pH}[/tex]
[H3O +] = [tex]10^{-1.2}[/tex]
We use the inverse of the log
[H3O +] = [tex]6.31*10^{-2}[/tex]
So the correct answer is [tex]6.31*10^{-2}[/tex]
