C3 TITRATION ----AQA!

THE CALCULATIONS:WHAT IS THE FORMULA???????

ANY STEPS FOR TACKLING A  TITRATION!?? WHAT I SHOULD DO FIRST?!!!!!!HELPPPP!

Posted Wed 23rd May, 2012 @ 20:30 by anonymous

EXAMPLE

KOH + HNO3 --> KNO3 + H2O

KOH = 20 grams and is 0.2mol/dm3

HNO3 = 15 grams and we want to work out it's concentration.

1 mole of KOH neutralizes 1 mole of HNO3 (as there are no numbers in front of the compounds in a balanced symbol equation)

No. of moles of KOH = concentration multiplied by volume

hence 0.2 * 20/1000 = 0.004 (this is the number of moles of KOH)

Because we know 1 mole of KOH neutralizes 1 mole of HNO3, we also know 0.004 moles of KOH will neutralize 0.004 moles of HNO3.

So using the equation we previously used to work out number of moles of KOH (moles = concentration * volume) we can find out the concentration of HNO3...

Moles = Concentration * Volume

so

0.004 = x (algebraic 'x') * 15/1000

or when rearranged...

0.004/(15/1000)

which equals 0.26666666667 but will round to 0.27mol/dm3

Hope this helps!

by the way, / is divided by and * is multiplied by, in case you were confused.

Also, remember to put units i.e. moles per decimetre cubed (mol/dm3)

and volume is as a fraction so 37cm3 (centimetres cubed) would be 37 over 1000 or 37/1000 as i have shown in the example.

Just watch out as well for numbers balancing the equation so if you have...

2x neutralising y, then y is twice as strong as x as it takes 2 of x to neutralise just 1 of y.

Good Luck!

Answered Wed 23rd May, 2012 @ 20:56 by Rick Abendstern

concentration = number of moles / volume

The volumes have to be in dm^3 but they usually give them to you in cm^3, so divide it by 1000. thats step one!

Answered Wed 23rd May, 2012 @ 20:46 by DennisTheMenace