standardization of NaOH solution:
a)volume of 6.0M NaOH solution used: 20mL
b)approximate[NaOH] after dilution to 400mL: 0.5M
c)weight of empty flask: 86.525g
d)weight of flask plus KHP: 87.253g
e)weight of KHP: 0.728g
f)moles of KHP: 0.00356mol
g)moles of NaOH: same answer as moles of KHP
h)initial buret reading: 0mL
I)final buret reading: 28.49mL
j)volume of NaOH used up: 28.49mL
k)[NaOH]: 0.00356mole/0.02849L NaOH= 0.12496M

Is it correct? Value of K is correct??

For b), if you used the 6.0 molar NaOH to get it, that would be 0.3 M

No. of moles of NaOH in 20 mL of 6.0 M NaOH = 0.12 mol
No. of moles of NaOH in the 400 mL diluted NaOH = 0.12 mol
No. of moles of NaOH in 1000 mL of diluted NaOH = (0.12/400) * 1000 = 0.3 mol

f) by KHP, you mean potassium hydrogen phosphide? I have never used this compound, but something is strange in that formula. K forms a +1 ion, H forms a +1 ion and P forms -3 ions, giving either KH2P or K2HP. If that's so, all your following work will carry the error.

g) If you took the Mr of KHP as (39.1+1+31) 71.1, then, you have (0.728/71.1) 0.0102 mol.

Btw, is that a practical that you performed or you were just given these values?

weight of KHP is 0.728g
so I think Moles of KHP is = mass/204.23g/mol
so, its 0.728g/204.23g/mol=0.00356mol

so, do you think that my k is correct?
[NaOHJ]=moles/L of NaOH=0.00356mol/0.02849L=0.12496M

in K do I have to use f) or b) ? I think I should use moles for f

please help me

I had to Google what KHP was...

Ok, yes, the answer is good.

However, you still haven't answered my above question,

Quote:

---

Originally Posted by me

is that a practical that you performed or you were just given these values?

---

Ohh.. sorry
Its my experimental values.
Do u think any value have problem?

pleaselet me know and Thank you so much

Yes, for b) I don't understand how you got 0.5M.

You have 0.12 mol of NaOH in 20 mL of 6.0M NaOH
You diluted that 0.12 mol from 20 mL to 400 mL. That gives a concentration of 0.3 M, not 0.5 M

You still have 0.12 mol in 400 mL.
400 mL -> 0.12 mol
1 mL -> 0.12/400 mol
1000 mL -> (0.12/400) * 1000 = 0.3 mol

So, [NaOH] = 0.3 M

Then, I don't know if they do it like that where you are, but I was told that we always had to make averages when titrating. I mean, perform minimum 2 titrations and use the average of the two volumes used.

okay thanks! For B
and yes I did test twice
but I just didn't wrote value of test 2

but l) I got 0.11421M
I added two tests[NaOH] and divided by 2.

Thanks

Oh, OK, things get clearer, phew! Ok, on to the next part, sample 2 now.