Question

A solution is prepared by adding 49.3 mL concentrated hydrochloric acid and 19.5 mL concentrated nitric acid to 300 mL water. More water is added until the final volume is 1.00 L. Calculate H [OH-), and the pH for this solution. [Hint: Concentrated HCl is 38% HCl (by mass) and has a density of 1.19 g/ mL; concentrated HNO3 is 70% HNO3 (by mass) and has a density of 1.42 g/mL.]
[H+] =
OH- =
pH =

Answer 1

Answer:

• [H⁺] = 0.920 M

• pH = 0.036

• [OH⁻] = 1.086x10⁻¹⁴ M

Explanation:

To calculate [H⁺] first we calculate the total number of H⁺ ions.

H⁺ from HCl ⇒ 49.3 mL * 1.19 gSolution/mL * 38 gHCl / 100 gSolution = 22.29 g HCl

• 22.29 g HCl ÷ 36.45 g/mol = 0.612 mol HCl = 0.612 mol H⁺

H⁺ from HNO₃ ⇒ 19.5 mL * 1.42 gSolution/mL * 70 gHCl / 100 gSolution = 19.38 g HNO₃

• 19.38 g HNO₃ ÷ 63 g/mol = 0.308 mol HNO₃ = 0.308 mol H⁺

Total H⁺ moles = 0.612 + 0.308 = 0.920 mol H⁺

Final Volume = 1.00 L

• [H⁺] = 0.920 mol / 1.0 L = 0.920 M

Now we calculate pH:

• pH = -log [H⁺]

• pH = -log(0.920) = 0.036

To calculate [OH⁻], we calculate pOH:

• pOH = 14 - pH

• pOH = 14 - 0.036 = 13.964

pOH = 13.964 = -log[OH⁻]

• $10^{-13.964}$ = [OH⁻] = 1.086x10⁻¹⁴ M