A solution prepared by dissolving 0.0084 moles of HCl in 1500 mL of solution has a Molarity of HCl of 5.6 × 10⁻³ M, a Molarity of H₃O⁺ of 5.6 × 10⁻³ M and a pH of 2.3.
A solution is prepared by dissolving 0.0084 moles of HCl in 1500 mL of solution. The molarity of HCl is:
[HCl] = 0.0084 moles / 1.5 L
= 5.6 × 10⁻³ M
HCl is a strong acid according to the following equation.
HCl(aq) + H₂O(l) ⇒ Cl⁻(aq) + H₃O⁺(aq)
Thus, the concentration of H₃O⁺ will be equal to the initial concentration of HCl, 5.6 × 10⁻³ M.
Now, we will calculate the pH of the solution using its Formula ;
pH = - log[H₃O⁺]
= - log [5.6 × 10⁻³]
= 2.3
Hence, A solution prepared by dissolving 0.0084 moles of HCl in 1500 mL of solution has a Molarity of HCl of 5.6 × 10⁻³ M, a Molarity of H₃O⁺ of 5.6 × 10⁻³ M and a pH of 2.3.
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