In assessing 20.0ml of NaOH 0.175M , calculate how many ml of 0.200M HCl must be added to obtain a pH of 12.55
millimols NaOH = mL*M = 20*0.175 = 3.5
If pH = 12.55, then pOH = 1.45 and (OH^-) = 0.03548 M so mmols at the end will be 0.03548*(20+xmL) where x mL is the volume of the 0.2M HCl added.
.....OH^- + H^+ ==> H2O
I...3.5.....0
add.......0.2x..........
C...-x...-0.2x.........s
E.0.03548*(20+x)
So we have mmols base-mmols acid = mmols base left over/total mL
3.5 - 0.2x = 0.03548(20+x) where is the mL of the 0.2 M HCl added.
Solve for x. The answer is close to 12 mL. You can check it this way.
You have 0.175 x 20 = mmols NaOH initially.
You add the value of mL HCl x 0.2 = mmols HCl added.
Subtract to find mmols NaOH left.
Then (OH^-) = mmols/total mL and convert to pH. I worked through it and it gives me 12.55 for pH. (pOH = 1.45)