Question

Soils and Hydrology

1a.) A corn crop requires about 120 lbs K/acre for a high yield, while a pine stand might need 25 lbs K/acre/yr to make good growth. Convert you exchangeable K value to lbs/acre and compare this to these two crop requirements for your soil. ( exchangeable K value is .09917 meq/100g)

1b.) A pratice titration problem: a 4g soil sample is extracted with 60 mL of NaCl solution, and 32 mL of this solution is titrated to the phenolphthalein endpoint with 6.2 mL of 0.005 N NaOH. Calculate the exchangeable acidity in meq/100 g.

Answer 1

Ans 1a) Corn crop K requirement = 120 lbs/acre

Exchangable K value = 0.09917 meq/100 gm or 0.09917 meq/0.22 lb or 0.45 meq/lb

=> Exchangable K requirement = 0.45 meq/lb x 120 lbs/acre = 54 meq/acre

We know, mg = meq x atmoic weight/ Valence

=> 54 meq = 54 x 39/1 = 2106 mg

=> Exchangable K requirement = 2106 mg/acre or 0.00464 lb/acre

Similarly,

Pine stand K requirement =25 lbs/acre

Exchangable K value = 0.09917 meq/100 gm or 0.09917 meq/0.22 lb or 0.45 meq/lb

=> Exchangable K requirement = 0.45 meq/lb x 25 lbs/acre = 11.25 meq/acre

We know, mg = meq x atmoic weight/ Valence

=> 11.25 meq = 11.25 x 39/1 = 438.75 mg

=> Exchangable K requirement = 438.75 mg/acre or 0.00096 lb/acre

Ans 1b) We know,

Exchangable acidity = (ml NaOH / W) x (0.005 mmol H / ml NaOH) x (0.1 cmol H / mmol H) x 1000 gm soil/kg

where, ml NaOH is volume of NaOH added

W = Mass of dry soil extracted with adding NaCl

Putting values,

=> Exchangable acidity = (6.2 ml / 4 gm) x (0.005 mmol / ml ) x 0.1 cmol/mmol x 1000 gm/kg

=> Exchangable acidity = 0.775 cmol/kg or 0.775 cmol / 1000 gm or 0.0775 cmol /100 gm or 0.775 mg/100 gm

NOTE : 1 cmol = 0.01 mol , so 0.0775 cmol is 0.000775 mol or 0.000775 gm or 0.775 mg

We know, meq = mg x Valance /atmoic weight

=> meq = mg x 1/1

=> Exchangable acidity = 0.775 mg/100 gm or 0.775 meq/100 gm