## Hwk 13.29

Edgar Khachatryan 3G
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am

### Hwk 13.29

1.436g of NaOH is dissolved in 300.mL solution, 25.00mL of this solution is titrated to the stoichiometric point with 34.20mL of 0.0695 M HCl. What is the percentage purity of the original sample?

1. I converted g of NaOH to moles of NaOH.
2. I then converted 0.0695M HCl to moles of HCl.
3. I then divided the moles of NaOH by .3L to get molarity of NaOH.
4. Then I multiplied molarity of NaOH to .025L to get moles of NaOH.
5. I subtracted moles of HCl from moles of NaOH.
6. I converted excess NaOH to grams.

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Hwk 13.29

dude ive been doing this problem for around an hour and i finally got it.
so first you find the moles of HCl (Because it's at the stoich point, and since NaOH is impure, we're gonna use HCl to find out NaOH)
0.0695 mol HCl x 0.03420 L = 0.0023769 moles HCl
so that means that there are 0.0023769 moles of NaOH
so we divide that by 0.25 L to find the molarity, and we get 0.0095076 M NaOH
then we multiply that molarity by 300 L to find the moles that were in the original sample, and we get 0.00286 moles
and then we multiply that by the MM of NaOH and get 0.11399 g rams
and then we divide that by 1.436 g to find the percentage of purity and we get 79.4% !!!

Edgar Khachatryan 3G
Posts: 20
Joined: Fri Jul 15, 2016 3:00 am

### Re: Hwk 13.29

Haha awesome, thank you!

Skolli20
Posts: 35
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Hwk 13.29

"dude ive been doing this problem for around an hour and i finally got it.
so first you find the moles of HCl (Because it's at the stoich point, and since NaOH is impure, we're gonna use HCl to find out NaOH)
0.0695 mol HCl x 0.03420 L = 0.0023769 moles HCl
so that means that there are 0.0023769 moles of NaOH
so we divide that by 0.25 L to find the molarity, and we get 0.0095076 M NaOH
then we multiply that molarity by 300 L to find the moles that were in the original sample, and we get 0.00286 moles
and then we multiply that by the MM of NaOH and get 0.11399 g rams
and then we divide that by 1.436 g to find the percentage of purity and we get 79.4% !!!"

This all looks good but shouldn't you be multiplying by 0.025 L? because its 25 ml in the titration?

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Hwk 13.29

you're right -- must have missed that. sorry! then all the calculations after that are off by a bit -- sorry again

Ryan Petrecca 1L
Posts: 16
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Hwk 13.29

Your final gram calculation for the NaOH is 1 decimal place off, but your final percentage is still correct...weird, must of just had a typo or something. *thumbs up*