15.0 ml of 0.50 M NaOH is added to a 100.-mL sample of 0.442 M NH3 (kb for NH3= 1.8 x 10^-5). What is the equilibrium concentration of NH4+ ions?
To find the equilibrium concentration of NH4+ ions, we need to set up an equation to represent the equilibrium reaction that occurs when NaOH reacts with NH3.
The balanced equation for the reaction is:
NH3 + NaOH → NH4+ + OH-
From the equation, we can see that for every mole of NaOH, one mole of NH4+ ions is formed. Therefore, the moles of NH4+ ions formed can be calculated using the stoichiometry of the reaction.
Step 1: Calculate the initial moles of NH3:
Initial moles of NH3 = concentration of NH3 × volume of NH3 solution
= 0.442 M × 100 mL
= 44.2 mmol (millimoles)
Step 2: Calculate the moles of NH4+ ions formed:
Since the moles of NH4+ ions formed are equal to the moles of NaOH used, we need to calculate the moles of NaOH first. This can be done using the concentration of NaOH and the volume of NaOH solution.
Moles of NaOH = concentration of NaOH × volume of NaOH solution
= 0.50 M × 15.0 mL
= 7.5 mmol
Since the reaction is 1:1 between NaOH and NH4+, the moles of NH4+ ions formed will also be 7.5 mmol.
Step 3: Calculate the final concentration of NH4+ ions:
Final concentration of NH4+ ions = moles of NH4+ ions formed / final volume of the solution
To calculate the final volume of the solution, we need to consider both the volume of NH3 solution and the volume of NaOH solution added.
Final volume = volume of NH3 solution + volume of NaOH solution
= 100 mL + 15 mL
= 115 mL
Final concentration of NH4+ ions = 7.5 mmol / 115 mL
Finally, we convert the concentration from mmol/mL to M (mol/L):
Final concentration of NH4+ ions = (7.5 mmol / 115 mL) × (1 L / 1000 mL)
= 0.0652 M
Therefore, the equilibrium concentration of NH4+ ions is 0.0652 M.