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MARUM - Sediment Geochemistry - Leobener Str - D-28359 Bremen - Germany

Marine Geochemistry - Laboratory Methods

Alkalinity titration
We use a fast titration to pH values well below pH 4.3 as described in Grasshoff(1983). Using this method, the pH needs to be determined as accurate as possible. Note that the formula for calculation of alkalinity in Grasshoff(1983) is incorrect. Using the equation below, the method is almost independent of both acid concentration and final pH-value allowing very rapid titration by adding a fixed acid volume and accurately determining the pH-value. We usually determine alkalinities from initial sample volumes as low as 1 ml(!)The correct equation calculates the alkalinity as buffered H+-ions per volume from the difference between absolute amount of H+-ions added (vHCl*cHCl) and the excess H+-ions after titration 10-pH*(v0+vHCl). Since the pH value is the log of the H+-ion activity this expression has to be divided by the activity coefficient of H+ (1 in pure water, approx. 0.8 in seawater, see Grasshoff(1983) for exact determination of fH+ or use PHREEQE)

Alkalinity (mol/l) = ( H+added - H+excess + H+ initial ) / v0

(1) Alkalinity (mol/l) = [(vHCl*cHCl) -10-pHfinal*(v0 +vHCl)/fH++ 10-pHstart * v0 /fH+] / v0

Examples

a)

pHstart = 8.02

vHCl = 1.32 ml (to pHfinal = 4.30)cHCl = 0.1 mol/l v0 = 50 ml fH+ = 0.8

Alkalinity = [(1.32/1000 * 0.1) – 10-4.3 * (50/1000 + 1.32/1000) / 0.8 + 10-8.02 * 50/1000 /0.8] / (50/1000)

Alkalinity = 0.00257 mol/l = 2.57 mmol/l

b)

pHstart = 4.95

vHCl = 22 ml ( pHfinal 3.912) cHCl = 0.001 mol/l v0 = 5 ml fH+ = 0.8

Alkalinity = [(22/1000 * 0.001) – 10-3.912 * (5/1000 + 22/1000) / 0.8+ 10-4.95 * 5/1000 /0.8] / (5/1000)

Alkalinity = 0.00358 mol/l = 3.58 mmol/l

 

If the initial pH is well above pH 5 the initial amount of H+-ions is neglectable and the equation simplifies to

Alkalinity (mol/l) = ( H+added - H+excess ) / v0

(2) Alkalinity (mol/l) = [(vHCl*cHCl) -10-pH*(v0+vHCl)/fH+] / v0

 

examples

pHstart = 8.02

vHCl = 1.32 ml (pHfinal = 4.30)cHCl = 0.1 mol/l v0 = 50 ml fH+ = 0.8

 

Alkalinity =[(1.32/1000 * 0.1) – 10-4.3 * (50/1000 + 1.32/1000) / 0.8 ] / (50/1000)

Alkalinity= 0.00257 mol/l = 2.57 mmol/l

 

b)

 but :

pHstart = 4.95

vHCl = 22 ml (pH final= 3.912)cHCl = 0.001 mol/l v0 = 5 ml fH+ = 0.8

 

Alkalinity = [(22/1000 * 0.001) – 10-3.912 * (5/1000 + 22/1000) / 0.8] / (5/1000)

Alkalinity = 0.00357 mol/l = 3.57 mmol/l

 

 

alkalinity calculation from different added volume / resulting pH value pairs of a real titration

 

advantages of this method:

exact even for small initial sample volumes and when acid concentration is low (large titration volume to initial sample volume ratio), Fast since the final pH only needs to be well below 4.3 - no exact pH-value has to be reached. With some experience acid has to be added only once or twice.

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