How on earth do you fill out this table?

titration.doc

ive tried everything possible to work out the moles or concentrations??

titration.docx191.6KB

Reply 1

I'd use a pen to fill it out. What have you tried? A carrot?

Reply 2

How many mol of H+ are there initially? What is the volume? What is the concentration of H+? Hence what is the pH?

Then add 10 cm3 of NaOH. How many mol of OH- is that? how many mol of H+ will the OH- react with? How many mol of H+ remain? What is the combined volume? What is the new concentration of H+? and hence pH?

Eventually, you'll have added enough OH- ions to remove all of the H+, after that point you'll start working out the number of mol of OH- in excess, you'll then still need the combined volume, but now the concentration of OH- ions and then (either using Kw or pOH) work out the pH.

Reply 3

TheBreatGritain

The first column you must fill in is very simply the vol NaOH in the already filled in column +25. eg 10+25=35, 15+25=40, 20+25=45

The next column requires you to calculate the conc of H+ ions. To do this you need to consider the reaction NaOH+HCL--->NaCl+H2O.
Here, you can simplify to OH-+H+--->H2O
So you need to calculate the moles of OH- from the NaOH (moles=conc*vol) then subtract this from the moles of H+ in HCL. The number then relates to the amount moles of H+ions you have. to calculate concentration, use conc=mol/vol taking into consideration the overall volume calculated in column 1
Using the equation pH=-log[H+] you then work out the pH by doing the -log of your answer.

However, if you have more OH- moles than H+moles you must find the moles of OH- remaining then work out the conc as above
You then you the ionic product of water equation. Kw=[OH-][H+]
to get Kw/[OH-]=[H+]
Kw= 1*10^-14.
You then put in the Kw value and [OH-] value to find [H+]
this you then put in pH=-log[H+]

it is a lot easier to explain in person ahaha

Reply 4

tariktarkan1907

OP

Original post by TheBreatGritain

The first column you must fill in is very simply the vol NaOH in the already filled in column +25. eg 10+25=35, 15+25=40, 20+25=45

The next column requires you to calculate the conc of H+ ions. To do this you need to consider the reaction NaOH+HCL--->NaCl+H2O.
Here, you can simplify to OH-+H+--->H2O
So you need to calculate the moles of OH- from the NaOH (moles=conc*vol) then subtract this from the moles of H+ in HCL. The number then relates to the amount moles of H+ions you have. to calculate concentration, use conc=mol/vol taking into consideration the overall volume calculated in column 1
Using the equation pH=-log[H+] you then work out the pH by doing the -log of your answer.

However, if you have more OH- moles than H+moles you must find the moles of OH- remaining then work out the conc as above
You then you the ionic product of water equation. Kw=[OH-][H+]
to get Kw/[OH-]=[H+]
Kw= 1*10^-14.
You then put in the Kw value and [OH-] value to find [H+]
this you then put in pH=-log[H+]

it is a lot easier to explain in person ahaha

youve been very helpful, so after the 25.01 cm3 volume of the hydroxide when you have more OH-, find the concentration of H+ using the ionic product equation right??
For the volumes below 25 of hydroxide, use the first method you described?

for the graph what ttwo variables do you use?

Reply 5

vol(OH-) on the X and pH on the Y.

Reply 6

tariktarkan1907

OP

Original post by Pigster

vol(OH-) on the X and pH on the Y.

ok thanks mate
so you have to use the Kw to work out the H+ when the hydroxide ion solution goes over 25cm3 right? LIKE 25.01 25.10 ETC

(edited 9 years ago)

Reply 7

Yes.

I teach the usage of pOH...

pOH = -log [OH-]

pH = 14 - pOH.

This only works at 25 oC, when Kw = 1.00 x 10^-14

What pigster said

OP

Ok i cant do the rest from 25.00cm3 when the OH- begins to increase more than the H+...the answers im getting dont make sense
When you use the Kw, you have to keep using the conc of the Oh- which is 0.2? o the values arent gonna be changing at all?

(edited 9 years ago)

Reply 10

n(HCl) initially = c x v = 0.20 x 25.00/1000 = 0.00500 mol

When v(NaOH) added = 25.01 cm3, then n(NaOH) added = c x v = 0.20 x 25.01 = 0.005002 mol

n(OH-) in XS = n(NaOH) added - n(HCl) initially = 0.005002 - 0.00500 = 0.000002 mol

Vtot = 25.00 + 25.01 = 50.01 cm3

[OH-] in XS = n / V = 0.000002 / (50.01/1000) = 3.9992 x 10-5 mol dm-3

(My way) pOH = - log [OH-] = 4.398
pH = 14.00 - pOH = 9.60

Reply 11

TheBreatGritain