Back titration Confused on the information given

[Zahta]

I really need help on a question off the heinemann chemistry text book on back titrations(volumetric analysis) on page 46 question 26.  They give you an equation  and  for the first section of the information they you give a standard solution of 20.00ml  if 0.1022 M of sodium hydroxide yet in the equation given there is no sodium hydroxide only hydroxide.  and later on they tell you its in excess how does that even make sense , then the excess sodium hydroxide is titrated with Hcl. i know i do two equations and stuff. But the information does not makes sense to me.


Thanks you anyone who helps, really appreciate it

[g1na]

The equation given is an ionic equation therefore sodium is not included for it is a spectator ion and also, there is an excess of sodium hydroxide to create a reaction with HCl.

i have scanned my working out for you as well

[Shenz0r]

It's important that you understand what's actually happening in a back titration. It might sound a little confusing at first, but let me give you an analogy.

Suppose that you're having a birthday party, you invite your friends, and you order a box of pizza (your initial amount of mol of the standard solution), which has eight slices of pizza. You and your friends start to eat the pizza. Then, you find out that two slices of pizza are remaining (which is how many mols of the standard solution is in excess). If you knew that one person could only eat one slice of pizza each (the molar ratio), then you can easily deduce that six slices were eaten and so six people must have been at the party (the amount of mol of your unknown substance). Hope that makes sense.

Work out the original number of mol of the standard solution, then work out the excess number of mol of it from your second titration. Remember that n(reacted) = n(initial) - n(excess), and then you'll have the number of mol of the standard solution that reacted with your unknown solution (you'll find out how much pizza was eaten at the party, and then how many people there were).

[Nobby]

Suppose that you're having a birthday party, you invite your friends, and you order a box of pizza (your initial amount of mol of the standard solution), which has eight slices of pizza. You and your friends start to eat the pizza. Then, you find out that two slices of pizza are remaining (which is how many mols of the standard solution is in excess). If you knew that one person could only eat one slice of pizza each (the molar ratio), then you can easily deduce that six slices were eaten and so six people must have been at the party (the amount of mol of your unknown substance). Hope that makes sense.

Hahaha that is quality

[ggxoxo]

It's important that you understand what's actually happening in a back titration. It might sound a little confusing at first, but let me give you an analogy.

Suppose that you're having a birthday party, you invite your friends, and you order a box of pizza (your initial amount of mol of the standard solution), which has eight slices of pizza. You and your friends start to eat the pizza. Then, you find out that two slices of pizza are remaining (which is how many mols of the standard solution is in excess). If you knew that one person could only eat one slice of pizza each (the molar ratio), then you can easily deduce that six slices were eaten and so six people must have been at the party (the amount of mol of your unknown substance). Hope that makes sense.

Work out the original number of mol of the standard solution, then work out the excess number of mol of it from your second titration. Remember that n(reacted) = n(initial) - n(excess), and then you'll have the number of mol of the standard solution that reacted with your unknown solution (you'll find out how much pizza was eaten at the party, and then how many people there were).

Great analogy!

[Zahta]

thank you guys so much