Question

what is the analyte concentration?

1L X 0.1M NaOH 39.997g NaOH 1 mol NaOH 100g Sol. 50g NaOH 1 ml 50 wt% 150 Sol 41 = 5.33 ml Sol. 1 mol NaOH Standardizing NaOH Trial Phthalate Mass (B) Rapid 0.5362g 0.5064g Initial Buret Final Buret Reading (ml) Reading (mL) 0.03 mL 28.6 mL 0.01 ml 25.8 ml Volume Added 28.3 mL Analyte Concentration 25.7 ml

Answer 1

SOLUTION:

RAPID:

We know that, Molar mass of Potassium hydrogen phthalate = 204.22 g/mol.

Given that, Mass of phthalate = 0.5362 g

Therefore, no.of moles of phthalate in the solution is given by,

n = given mass 0.5362 g molar mass204.22 g/mol

:in = 0.0026256 mol

Given that, Volume added, V = 28.3 mL = 28.3 x 10^-3 L

Therefore, Concentration is given by,

n 0.0026256 mol C= v= 28.3 x 10-3 L

::C= 0.0928 M 0.1 M

TRIAL 1:

Given that, Mass of phthalate = 0.5064 g

Therefore, no.of moles of phthalate in the solution is given by,

n = given mass 0.50649 molar mass204.22 g/mol

:in = 0.0024797 mol

Given that, Volume added, V = 25.7 mL = 25.7 x 10^-3 L

Therefore, Concentration is given by,

n 0.0024797 mol C = v= 25.7 x 10-3 L

::C= 0.0965 M 0.1 M