**What quantity (moles) of NaOH must be added to one liter of 2M HC2H3O2 to produce a solution buffered at each pH:**

A) pKa = pH (I figured this one out no problem)

B) pH=4 (answer is 0.3 mol)

C) pH=5 (answer is 1.3 mol)

Thanks for the help!

Gold3nLily

Oh yes and the ka= 1.8 x 10^-5

for b...

ph= pka + log (a-/ha)

4=4.75 + log a-/ha

-0.75 = log a-/ha

10^-0.75 = a-/ha

0.177827941 = a-/ha

Technically in the A-/HA it says to write it as molarity, but it you write it as molarity say 1 mol A-/ L / 2 mol HA / L the liters cancel out since both will be at the same volume in the end. So we can just write it as moles.

So 0.177827941 = a-/ha

If we have 2 mols of HA (problem says 2 M and 1 liter so 2 moles) and we add NaOH, the base NaOH will react with the acid HA. So however many moles of OH- we add is how many moles of HA we lose and how many moles of A- we gain.

So we can write it as

0.177827941 = x moles A- / 2-x moles HA

Solve for x and you get 0.30196 moles.

same work for c except change the ph.