How would you calculate the surface tension of a 2% (w/v) solution of a wetting agent that has a density of 1.008 g/mL and that rises 6.6 cm in a capillary tube having an inside radius of 0.2 mm?

1 Answer
Aug 30, 2016

Answer:

The surface tension is #"73 mJ·m"^"-2"#.

Explanation:

One method to measure the surface tension of a liquid is to measure the height the liquid rises in a capillary tube.

www.chem1.com

The formula is

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/"2cosθ" color(white)(a/a)|)))" "#

where

#γ# = the surface tension
#r# = the radius of the capillary
#h# = the height the liquid rises in the capillary
#ρ# = the density of the liquid
#g# = the acceleration due to gravity
#θ# = the angle of contact with the surface

For pure water and clean glass, the contact angle is nearly zero.

If #θ ≈ 0#, then #cosθ ≈ 1#, and the equation reduces to

#color(blue)(|bar(ul(color(white)(a/a) γ = "rhρg"/2 color(white)(a/a)|)))" "#

In your problem,

#r = "0.2 mm" = 2 × 10^"-4"color(white)(l) "m"#
#h = "6.6 cm" = "0.066 m"#
#ρ = "1.008 g/mL" = "1.008 kg/L" = "1008 kg·m"^"-3"#
#g = "9.81 m·s"^"-2"#

#γ = (2 × 10^"-4" color(red)(cancel(color(black)("m"))) × 0.066 color(red)(cancel(color(black)("m"))) × 1008 color(red)(cancel(color(black)("kg·m"^"-3"))) × 9.81 color(red)(cancel(color(black)("m·s"^"-2"))))/2 × ("1 J")/(1 color(red)(cancel(color(black)("kg")))·"m"^2color(red)(cancel(color(black)("s"^"-2"))))= "0.065 J·m"^"-2" = "65 mJ·m"^"-2" #

For comparison, the surface tension of pure water at 20 °C is #"73 mJ·m"^"-2"#.