# Accounting for DI Water during Titration

1. Jun 26, 2013

### giacomh

So I've gotten close to the correct answer for this problem, but I think I'm accounting for the DI water incorrectly:

In titrating a blank, 0.11 mL of sodium hydroxide solution was required to titrate 50.00 mL of DI water. It required 17.56 mL of the same sodium hydroxide solution to titrate 0.7711 g of potassium hydrogen phtalate (KHC8H4O4) dissolved in 20.00 mL of D water. Calculate the molarity of the NaOH solution.

I accounted for the DI water by simply subtracting the NaOH required for the DI titration from the NaOH required for the KHP titration. I think I need to account for the change in volume? Not sure how to do that...

My attempt:

17.56mL - .11 mL = .1745 mL
.7711 g / 204.2234 g/mol = .003776 mol
.003776 mol/ .01745 L = .2164 M

Thanks!

2. Jun 27, 2013

### Staff: Mentor

Hard to say.

Naive approach would call for assuming if 50 mL required 0.11 mL for a blank, 20 mL will require 20/50*0.11 mL. But it doesn't have to be true, as blank can require a constant volume (for example used to change color of the indicator) and/or volume of titrant that depends on the volume of the sample. In lab reality you should dissolve KHP in 50 mL, then it would not matter what the blank is really needed for. (Actually in lab reality I would dissolve KHP separately to be able to repeat the titration, but that's another story).

As the question is poorly designed, there is no good answer here. What you did is what I would do, whether it is what the person asking the question had on mind I have no idea.