A student dissolves an unknown mass of sodium hydroxide in water to make 200cm3 of an aqueous solution. A 25cm3 sample of this sodium hydroxide is placed in a conical flask and is tut rated with 0.015 moldm-3 sulfuric acid. Titre = 19.58
Calculate mass of sodium hydroxide used to make original solution
2NaOH + h2so4 -> na2so4 + 2h20 For this question I got 2.4g, can someone try this question and see what they get
The first thing you are meant to do for this question, is to calculate the mean titre. As the values 19.55 and 19.60 from the table are concordant, you use the two values to calculate the mean titre. 19.55 + 19.60/2 = 19.58cm3
Hi, for this question I got 0.188g. Moles of sulphuric acid = 19.58cm3 x 0.015moldm-3 / 1000= 0.0002937mol Ratio of NaOH to H2SO4 = 2:1 therefore the moles of NaOH = 0.0002937 x 2 = 0.000587mol That was for the 25cm3 sample so for the 200cm3 sample the moles of NaOH would be: 0.000587 x 8 = 0.0047mol. Mr of NaOH + 23 + 16 + 1 = 40 Mass = moles x Mr = 0.0047 x 40 = 0.188g
Your conc for this question are wrong. I know this was 2 years ago but it's can be confusing for anyone seeing this later. The correct concentration is 0.15 mol dm -3 for h2so4