Times the 100 milliliters, or 0.1 liter,

and that gives us 0.0750 moles.

So I'll put 0.0750 there.

So the smallest is going to be reacting, so

I will have none of the hydroxide, it has been neutralized as a buffer will do.

We will subtract 0.0003 from here,

and add 0.0003 to this value.

That is going to give to us, in the F line

under ammonium, 0.0745 and

for ammonia, NH3, 0.0753.

Okay at this point we have buffer so we can determine the pH of the buffer.

pH = -log of the KA of ammonium,

so that's 1.0 x 10 to the -14,

divided by 1.76 x 10 to the -5.

That's the pKa, that whole portion here is the pKa,

plus the log, and we can use moles,

the moles of the base is on top, so that's 0.0753,

over the moles of the acid, 0.0745.

So this will give a pH equal to this first

term equals to 9.246 plus the second term,

which is 0.0046 for

final pH to three decimal places of 9.250