Solution mixing 25.0 mL of 0.250 M nitric acid (HNO3) with 15.0 mL of 0.500 M sodium hydroxide (NaOH). Final concentration of each ion?
A solution is made by mixing 25.0 mL of 0.250 M nitric acid (HNO3) with 15.0 mL of 0.500 M sodium hydroxide (NaOH). What is the resulting concentration of each ion (H+, NO3-, Na+, OH-) in solution?
The answers are in the table inthe picture attached. Could someone please explain the reasoning and steps to find the answers on the right side of the table? The Concentrations AFTER reaction is completed?
- hcbiochemLv 71 month ago
For Na+ and NO3- which are spectator ions, simply use the formula:
M1V1 = M2V2
For Na+: 0.500 M (15.0 mL) = M2 (40.0 mL)
M2 = 0.188 M
0.250 (25) = M2 (40)
M2 = 0.156 M
For H+ and OH-, these react to form water, with one of those ions being in excess. So, first calculate moles of each available:
moles H+ = 0.0250 L X 0.250 mol/L = 6.25X10^-3 mol H+
moles OH- = 0.0150 L X 0.500 mol/L = 7.5X10^-3 mol OH-
H+ is the limiting reactant, leaving 1.25X10^-3 mol OH-
Molarity OH- = 1.25X10^-3 mol / 0.040 L = 0.0313 M
Molarity H+ = 1.00X10^-14 / 0.03133.2X10^-13 M