A supply of NaOH contains the contaminants NaCl and MgCl2. A 4.9995g sample of this material is dissolved and diluted to 500mL, 20mL of this is titrated with 22.26mL of a 0.1989M solution of HCl. What is the percent NaOH of the original solution?

1 Answer
il maestro
May 6, 2015

88.5%

22.26 ml of 0.1989M HCl contain #0.02226*0.1989= 4.427*10^-3# moles of HCl. Each mole of HCl titrate 1 mole of NaOH. If 20 ml of the solution contains #4.427*10^-3# moles, in 500 ml there were 0.1107 moles of NaOH. The weight of NaOH in the sample is #0.1107*40(MM of NaOH)=4.4275 # g
4.9995 g of crude material contain 4.4275 g of NaOH, therefore #4.4275/4.9995*100= 88.55%#

Related questions
Impact of this question
3541 views around the world
You can reuse this answer
Creative Commons License