Titration calculation PDF

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The best way to do a titration using an ICE table....

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Titration calculation – how to example A 0.1950 M solution of HCl was titrate with an unkown concentration of NaOH. When the titration was performed, 18.00mL of the HCl solution was required to neutralise 25.00mL of the NaOH. Determine the concentration of NaOH. 1. Write down an equation for the reaction HCL (aq) + NAOH(aq)  H2O(l) + Na+(aq) + Cl-(aq) 2. Identify the known and unknown reagents n=c x v n (moles) c (concentration) v (volume)

HCl

NaOH

0.1950 M 0.0180 L

? 0.0250 L

From the table we can see that we know two of the three terms for the HCl, so this is our known reactant.

3. Calculate moles for known reactant (n) HCl = (c) HCl x (v) HCl So, (n) HCl = 0.1950 M x 0.0180 L HCl = 0.003510 moles. Add to table n=c x v n (moles) c (concentration) v (volume)

HCl 0.003510 0.1950 M 0.0180 L

NaOH ? 0.0250 L

4. Identify the mole ratio from the equation and calculate the moles of your unknown compound H+(aq) + OH-  H20(l) We have a 1:1:1 ratio, so moles of our unknown = moles known Therefore (n) = NaOH = 0.003510 Add to table n=c x v n (moles) c (concentration) v (volume) 5.

HCl 0.003510 0.1950 M 0.0180 L

NaOH 0.003510 ? 0.0250 L

Calculate the concentration of your unknown reagent

We can see in our table that we have 2 of three terms for our unknown reagent (NaOH), so we can calculate the third. Rearrange equation n = n/v So 0.003510 M/ 0.0250 L =0.1404 M...