Problem: A solution is prepared by adding 50.0 mL concentrated hydrochloric acid and 20.0 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.]

FREE Expert Solution

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

HCl and HNO3strong acids → completely ionize in the solution


Calculate moles Hfrom HCl:

HCl → H+ + Cl-

38% HCl (by mass) = 38 g HCl/100 g solution

molar mass HCl = 36.458 g/mol

mass HCl=50.0 mL solution×1.19 g solution1 mL solution×38 g HCl100 g solution

mass HCl = 22.61 g HCl

moles H+=22.61 g HCl×1 mol HCl36.458 g HCl×1 mol H+1 mol HCl

moles H+ = 0.6201 mol


Calculate moles Hfrom HNO3:

HNO3 → H+ + NO3-

70% HNO3 (by mass) = 70 g HNO3/100 g solution

molar mass HNO3 = 63.018 g/mol

mass HNO3=20.0 mL solution×1.42 g solution1 mL solution×70 g HNO3100 g solution

mass HNO3 = 19.88 g HNO3

moles H+=19.88 g HNO3×1 mol HNO363.018 g HNO3×1 mol H+1 mol HNO30.

moles H+ = 0.3154 mol


Total moles H+0.6202 mol + 0.3154 mol

Total moles H+ = 0.9356 mol


Calculate [H+]:

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Problem Details

A solution is prepared by adding 50.0 mL concentrated hydrochloric acid and 20.0 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.]

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