In: Chemistry
A mixer used to prepare 1000 kg of 8 wt% sodium hydroxide (NaOH) solution. The feeds to the mixer are 10 mol% sodium hydroxide solution in water and pure water. What are the quantities (in kg) of pure water and 10 mol% sodium hydroxide solution to be added to the tank?
Initially we have 10 mol % NaOH which means 10 moles NaOH per 100 mol solution
water moles = 100-10 = 90
mass of NaOH = moles x molar mass of NaoH = 10 x 40 g/mol = 400 g
water mass= 90 x 18.015 = 1620 g
Thus for 400 g NaOH we get 1620 g water in 10 % NaOH solution or to say 400 kg NaOH per 1620 kg water
we need 8 wt % NaOH which means 8 kg NaOH per 100 kg solution
we need 1000 kg such solution
Hence NaOH needed = ( wt % of NaoH / 100) x mass of solution
= ( 8/100) x 1000 = 80 kg
Thus we need 80 Kg NaOH and water mass of 1000-80 = 920 kg
when we use 10 % mol NaOH water mass we get is 1620 kg per 400 kg NaOH
water mass = ( 1620 /400) x 80 = 324 kg
Thus from 10 mol % NaoH we get water mass of 324 kg
pure water required = total water required - water from 10 mol % NaoH
= 920-324 = 596 kg
Mass of 10 mol % NaoH = NaoH mass + water mass from it = 80+324 = 404 kg
Thus to get 1000 kg of 8 % wt NaOH we need 404 kg of 10 mol % NaOH and 596 kg of pure water