A solution is prepared by adding 47.3 mL of concentrated hydrochloric acid and 16.3 mL of concentrated nitric acid to 300 mL of 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 and has a density of 1.42 g/mL.]
2 answers
Find [H+] [OH-] and Ph
The easiest way to explain this (but perhaps the long way around) is to first convert to molarity for both HCl and HNO3.
For HCl:
1.19 g/mL x 1000 mL x 0.38 x (1 mol/molar mass HCl) = mols HCl/L = M
For HNO3:
density g/mL x 1000 mL x 0.70 x (1 mol/molar mass HNO3) = moles HNO3/L = M
Then mols HCl = M x L = M x 0.0473 = ?
mols HNO3 = M x 0.0163 = ?
Add mols HCl to mol HNO3.
M soln = total mols/total volume (which is 1.0 L final volume).
Note: Those aren't hints at the end of the problem. That information is essential to the problem.)
For HCl:
1.19 g/mL x 1000 mL x 0.38 x (1 mol/molar mass HCl) = mols HCl/L = M
For HNO3:
density g/mL x 1000 mL x 0.70 x (1 mol/molar mass HNO3) = moles HNO3/L = M
Then mols HCl = M x L = M x 0.0473 = ?
mols HNO3 = M x 0.0163 = ?
Add mols HCl to mol HNO3.
M soln = total mols/total volume (which is 1.0 L final volume).
Note: Those aren't hints at the end of the problem. That information is essential to the problem.)
2 answers