Full correction in mmol = 0.5 x Base deficit (mmol/l) x Wt. (kg)
This is an approximation based on an estimated volume of distribution of bicarbonate of 0.5 litres/kg body weight.
Base deficit = 24 - Bicarbonate concentration
The bicarbonate concentration is 14 mmo/l.
Full correction = 0.5 x 10 x 15 = 75 mmol
Half correction = 37.5 mmol
8.4% NaHCO3 contains 1 mmol/ml of bicarbonate.
This is very hyperosmolar and should be diluted as per guidelines.
Treatment of hyperkalaemia is discussed in the Management of Fluid and Electrolyte Problems in Children module.
The simplest option is to give nebulized salbutamol.
Correction of the acidosis should also help shift potassium into cells
Removal of potassium can be achieved used calcium resonium either oraly or rectally.
Continued problems with hyperkalaemia would be an indication for dialysis.
You can now go back to bed.