Please calculate the pH 0f an aqueous solution containing 1mM NaOH containing 1mM NaOH plus C M NH4OH?

Question

(pKb=4.76) where C=1 and .01 M

Answer 1

NaOH is a strong electrolyte. thus dissociation is complete and you have [OH-]=10^-3 M. Therefore pOH=-log[OH-]= 3.00 and pH=14-pOH=> pH= 11.00.

In the second case you have a weak base and a strong base at the same time. So assuming that we have the OH- from the strong base as initial amount (10^-3M) let's see how the dissociation of the weak electrolyte will be affected

.. .. .. .. .. .. NH4OH <=> NH4(+) + OH(-)
Initial. .. .. .. .. .. C .. .. .. .. .. .. .. .. .. .. 10^-3
Dissociate .. .. . x
Produce .. .. .. .. x .. .. .. .. .. x.. .. .. .. .. x
At Equilibrium.. C-x .. .. .. .. .x .. .. .. .. 10^-3 +x

Kb= [NH4+][OH-] / [NH4OH] = x (10^-3 +x)/(C-x) =>
CKb - xKb= x10^-3+ x^2 =>
x^2 + (Kb+10^-3)x -KbC =0
Thus
x=0.5*(-(Kb+10^-3)) + squareroot( (Kb+10^-3)^2 + 4KbC)
Kb=10^-Kb=10^-4.76 therefore

x=0.5*( -(10^-4.76+10^-3) +SQRT( (10^-4.76+10^-3)^2 +4*(10^-4.76)*C))

For C=1 M
x=3.69*10^-3
Thus pOH=-log (10^-3+x)= -log(10^-3+3.69*10^-3)= 2.33
and pH= 14-2.33 =11.67

For C=0.01 M
x=1.49* 10^-4
pOH= -log(10^-3+ 1.49*10^-4)= 2.94
and pH=14-2.94= 11.06

Answer 2

If pH = 7.6, then pOH = 14-7.6 = 6.4 Because the pH/pOH is very close to 7.00, it is insufficient to calculate: [OH-] = 10^-pOH = 10^-6.4 = 3.98*10^-7 as being the molarity of the NaOH Water dissociates to produce [OH-] = 10^-7, so this [OH-] must be subtracted from the total [OH-] in order to isolate the [OH-] originating from the NaOH [OH-] from NaOH = (3.98*10^-7) - (10^-7) = 2.98*10^-7 The concentration of the NaOH in solution = 2.98*10^-7M

Answer 3

can't calculate, but it sounds like 7