What is the ph of 0.11 m diethylamine, (ch3ch2)2nh, (kb = 8.6 × 10−4)?

Answer : The pH of the solution is, 11.97

Solution :  Given,

Concentration (c) = 0.11 M

Base dissociation constant =

The equilibrium reaction for dissociation of (weak base) is,

initially conc.         0.11                              0             0

At eqm.              (0.11-x)                            x             x 

First we have to calculate the value of 'x'.

Formula used :

Now put all the given values in this formula ,we get:

By solving the terms, we get

Thus, the concentration of hydroxide ion is:

Now we have to calculate the pOH.

Now we have to calculate the pH.

Therefore, the pH of the solution is, 11.97

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

the answer is c. nacl