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.
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.
Why do you divide .01 by 2. Shouldn't you divide by 1 because 50 ml+50ml=1L
To find the time required for the OH- concentration to reach 1e-4 M, we need to use the rate equation for a first-order reaction. The rate equation for a first-order reaction is given by:
(rate) = k[A]
Where:
(rate) is the rate of the reaction
k is the rate constant
[A] is the concentration of the reactant A
In this case, the reactant A is OH-.
We are given that the reaction is kinetically first order in both OH- and NH4+ concentrations, and the rate constant (k) is given as 2e10 1/M s.
To solve this problem, we first need to determine the initial concentration of OH-. Given that 50 ml of 0.01 M NaOH (sodium hydroxide) is mixed with 50 ml of 0.01 M NH4Cl (ammonium chloride), we can assume that the total volume of the mixture is 100 ml or 0.1 L. Since NaOH and NH4Cl are completely dissociated in water, the initial concentration of OH- is equal to the concentration of NaOH, which is 0.01 M.
Now we can use the rate equation to find the time required for the OH- concentration to reach 1e-4 M. Rearranging the equation, we have:
(rate) = k[OH-]
t = 1 / (k[OH-])
Substituting the values, we get:
t = 1 / (2e10 * 1/M * 1e-4 M)
t = 1 / (2e10 * 1e-4)
t = 1 / (2e6)
t = 5e-7 s
Therefore, the time required for the OH- concentration to reach 1e-4 M is 5e-7 seconds.