How many grams of sodium borate MW 201 are required to prep

How many grams of sodium borate (MW = 201) are required to prepare 1 liter of solution containing 0.4 M of boric acid and having a pH of 8.9? The pKa of boric acid is 9.24?

Solution

molarity = no of moles of solute in 1 litre of solution

given molarity = 0.4 ;

M = no of moles / 1 litre of solution

where no of moles = weight / gram molecular weight

no of moles = molarity x 1 litre of solution = 0.4;

weight = no of moles X gram molecular weight = 0.4 X 201 = 80.4 gm

80.4 grams of sodium boarate is required to prepare 1 litre of solution containing 0.4M of boric acid.