The reaction OH- + NH4+ = H20 + NH3 is kinetically first order in both reactants' concentrations. The rate constant is 2e10 1/M s. If 50 ml of .01 M NaOH is mixed with 50 ml of .01 M NH4CL, find the time required for the OH- concentration to reach 1e-4 M.

Your answer: It starts at 0.01/2 = 0.005 and you want it to go to 1E-4
ln(5E-3/1E-4) = 2E10*t
Solve for t.

My new question is: Why do you divide .01 by 2. Shouldn't you divide by 1 because 50ml+50ml=1L

Six of one but all roads lead to the same answer.

You start with 0.01M NaOH and you dilute it from 50 to 100 which is a factor of 2.
0.01/2 = 0.005M OR
0.01M x (100/200) = 0.005 OR
mols NaOH = M x L = 0.01M x 0.05L = 0.0005 mols and M = mols/L = 0.0005/0.1L = 0.005M (50 mL + 50 mL = 100 mL = 0.1L)
Voila!

In the given reaction, OH- + NH4+ = H2O + NH3, the stoichiometric coefficient between OH- and NH4+ is 1:1. Therefore, when equal volumes of 50 mL of 0.01 M NaOH and 50 mL of 0.01 M NH4Cl are mixed, their concentrations become equal as well.

To calculate the initial concentration of OH- after mixing, we first need to find the total volume. Since the volumes are additive, 50 mL + 50 mL = 100 mL = 0.1 L.

Next, we calculate the total number of moles of OH-. The molarity (M) is defined as moles per liter, so the number of moles of OH- is given by:

moles of OH- = volume of OH- solution (L) * molarity of OH- (mol/L)
= 0.1 L * 0.01 mol/L
= 0.001 mol

Since the stoichiometric coefficient is 1:1, the number of moles of NH4+ will be the same as that of OH-. Therefore, the total number of moles of OH- and NH4+ after mixing is 0.001 mol + 0.001 mol = 0.002 mol.

Now, we divide the total number of moles (0.002 mol) by the total volume (0.1 L) to obtain the concentration of OH- after mixing:

concentration of OH- = moles of OH- / total volume
= 0.002 mol / 0.1 L
= 0.02 M

Therefore, the initial concentration of OH- after mixing is 0.02 M. However, it's important to note that this concentration is double the initial concentration of NaOH (0.01 M) or NH4Cl (0.01 M) individually. Hence, we divide by 2 to obtain the starting concentration of OH-.

To answer your question, the division by 2 reflects the fact that when equal volumes of NaOH and NH4Cl are mixed, their concentrations become equal, and thus the initial concentration of OH- is half the individual concentration of NaOH or NH4Cl. This is taken into account in order to accurately calculate the initial concentration of OH-.