Question

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?

Answer 1

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