

InterviewSolution
Saved Bookmarks
1. |
Calculate the volume of `1.00 mol L^(-1)` aqueous sodium hydroxide that is neutralized by `200 mL` of `2.00 mol L^(-1)` aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is, `NaOH_((aq.))+HCl_((aq.))rarr NaCl_((aq.))+H_(2)O_((l))` |
Answer» Step I. Calculation of volume of NaOH solution neutralised Volume of NaOH solution neutralised can be calculated by applying molarity equation `M_(1)V_(1)=M_(2)V_(2)` `(1 M)xxV_(1) = (2M)xx(200 mL)=400 mL` Step II. Calculation of mass of `NaCl` produced The neutralised reaction is : `NaOH(aq)+underset(underset(36.5g)(("1 mol")))(HCl(aq))rarrunderset(underset(58.5g)(("1 mol")))(NaCl(aq))+H_(2)O(l)` `:.` Mass of HCl in 200 ml of 2M solution can be calculated as follows : molarity of solution `= ("Mass of HCl/Molar mass")/("Volume of solution in litres")` `(2.0"mol L"^(-1))=("Mass of HCl")/(("36.5 g mol"^(-1))xx(0.2L))` Mass of HCl `= (2.0 "mol L"^(-1))xx("36.5 g mol"^(-1))xx(0.2 L)=14.6 g` Now, `36.5 g` of HCl produce NaCl after neutralisation = 58.5 g `:.` 14.6 of HCl produce NaCl after neutralisation `= ((58.5g))/((36.5g))xx(14.6g)=23.4g`. |
|