Titration Calculations Made Easy Chem GCSE Tomorrow! Please Help Me!?

Question

I have no clue, so any step by step information would be fantastic, thanks in advance! x

Answer 1

to start with, you need to have a balanced equation otherwise the ratios will be all out!I find it easiest to put all the information in a little table and then work out what information you have and where it fits into the formulas Moles=concentration x volume and moles=mass/relative formula or atomic mass.

once you have the balanced equation, you need to identify the ratio of the substances involved so, for example 2HCl and NaOH would be in the ratio 2:1.

Put all the information such as concentration, volume (remember it must be in dm cubed), mass etc. and work out how many moles of one substance there are and then from the ratio you can work out the number of moles in the other substance.

you can then work out the concentration of the substance from all that information.

for example:

calculate the concentration of HCl when 22cms³ of it are needed to neutralise 25cms³ of 0.5M NaOH:

balanced equation: HCl + NaOH → NaCl + H2O

                                    HCl                                    NaOH

conc.                            ?                                        0.5M

volume:                     22cms³                             25cms³

                                 0.022dm³                          0.025dm³

moles=conc x vol     ?                                        0.5 x 0.025

                                                                               0.0125

conc=moles/volume  0.0125/0.022

                                    0.568M

concentration of HCl = 0.568Molar

Hope this helps!!

Answer 2

are you doing the AQA additional, if so i am doing the sam exam and can help you

Answer 3

right

im learning this now too :)

so basically.. you want to figure out the concentration of one of the solutions

25 cm^3 (cubed) of sodium hydroxide (unknown concentration)

reacts with exactly 20 cm ^3 of 0.5 mol/dm^3 hydrochloric acid

step one... look at the chemical formula to work out how many moles of each solution react with the othe

chemical equation : NaOH(aq) + HCL (aq) -----> NaCL + H2O

so 1 mole of NaOH

reacts with 1 mole of HCL

we know the concentration of the HCL is 0.5 mol/dm^3

and a decimeter = 1000 cm^3

so 0.5/1000 moles of HCL are dissolved in 1cm^3 of acid

now scale it up to the amount you actually have...so thats 20 cm^3 of HCL

0.5/1000 X 20 = 0.01 moles of HCL dissolved in 20cm^3

and because 1 mole reacts with 1 mole

0.01 moles reacts with 0.01 moles

there must be 0.01 moles in the 25cm^3 of sodium hydroxide you have

and in 1dm^3 (1000cm^3)

0.010/25 of NaOH dissolved in 1cm^3 of solution

and there are 1000 in a dm^3

0.010/25X1000=0.4 mol/dm^3 (concentration of solution)

its BASICALLY from the book

but i guess im typing it up to help me too ;)

good luck xx