# Find the wt% KHP in an unknown via titration?

Submitted by nicole.marie on Tue, 09/30/2014 - 22:30

So in lab we had to find the wt% of KHP in an unknown (with another soluble salt). The wt% varies between 15-80%.

We used 0.1 M NaOH.

I had 10.0216 g of unknown disolved in 250 mL of H2O.

40 mL of the KHP solution took 29.90 mL NaOH and 29.85 mL NaOH to neutralize.

To find the wt%, this is my work

0.02990 L NaOH x 0.10mol/L NaOH = 0.002990 moles (of NaOH and KHP)

0.002990moles / 0.040 L = 0.07475m/L of KHP

0.07475moles x 204.218 g/mole = 15.265 g KHP / L

but because this was in 40 mL, I took 15.265 g x 0.04 = 0.610612 g/40 mL

and since it stared with 250 mL, I multiplied 0.6106 x (250/40) and got 3.81625 grams

3.816/10.0216 g  x  100 = 38.063 wt%

(and did the same for 29.85)

I'm just not sure if this is the correct way to do this. No one else that I've talked to has done it this way.

Any help would be appreciated!

0.002990 moles (of KHP) as 1 mole NaOH reacts with 0.00299 moles KHP. This was in 40mL, so 250 mL would contain 0.00299 x (250/40) moles of KHP = 0.0186875 moles in 250 mL solution. This will have a mass of 0.0186875 x  204.218 g/mole = 3.816323875g and the percentage would be (3.816323875/10.0216) x 100