If you mix two solutions, 24.0 mL of 1.498 M NaCl and 26.0 mL of 1.995 M NaOH....?
If you mix two solutions, 24.0 mL of 1.498 M NaCl and 26.0 mL of 1.995 M NaOH, calculate the final sodium (Na+ ), chloride (Cl - ) and hydroxide (OH- ) ion concentrations.
- pisgahchemistLv 72 weeks ago
Mixing ions ....
NaCl and NaOH ionize completely.
Na+ .... Cl- .... OH-
[Na+] = (0.0240L x (1.498 mol Na+ / 1L) + 0.026L x (1.995 mol Na+ / 1L)) / 0.0500L = 1.76M Na+
[Cl-] = 0.0240L x (1.498 mol Na+ / 1L) / 0.0500L = 0.719M Cl-
[OH-] = 0.0260L x (1.995 mol Na+ / 1L) / 0.0500L = 1.04M OH-
The answers can only be expressed to three significant digits.
- Bobby_ThinLv 72 weeks ago
when these two solutions are mixed there is no reaction so we calculate the molarities separately in 50mL of combined volume
moles of Cl- = 1.498 mol /L * 0.0240L = 0.0360 mole
in the new volume of of 50 mL molarity = 0.0360 moles / 0.050L = 0.719 M
moles of OH- = 1.995 moles / liter * 0.026 L = 0.0519 mole
in 50.0 mL = 0.0519 moles / 0.050L = 1.037 M
Moles of Na+ = (0.0360 mole + 0.0519 mole) / 0.050L = 1.76 M