Emil Heinäaho

CONCENTRATIONS AND MOLES

The calculations for this work are the following.

1A. Calculate the volume of HCl you need to get 100 ml of dilute acid.

The density of HCl = 1,19 kgdm^-3 ( 37%) = 1190 gdm^-3

n_HCl = m/M = 1190g / (1,01+35,45) gmol^-1 = 34,64 mol

c_HCl = 32,64 mol * 37% = 12,08 moldm^-3

n_HCl = c_HCl * v_HCl = 0,2 moldm^-3 * 0,1dm³ = 0,02 mol

v_HCl = n_HCl / c_HCl = 0,02mol / 12,08moldm^-3 = 0,0017dm³ = 1,7ml

2A. HCl (aq.) + NaOH (aq.) = Na⁺(aq.) + Cl^-(aq.) + H₂O (l.)

c_NaOH = 0,1 moldm^-3

v_NaOH = measured

n_NaOH = measured

v_HCl = 20 ml

n_NaOH = n_HCl

Next thing is to calculate the concentrate of HCl.

NaOH

C_NaOH = 0,1 moldm^-3

V_NaOH = 0,034 dm³

N_NaOH = ?

n_NaOH = 0,1 moldm^-3 * 0,034 dm³ = 0,0034 mol

HCl

v_HCl = 0,02dm³

n_HCl = 0,0034 mol (n_NaOH = n_NaOH !)

c_HCl = 0,0034 mol / 0,02dm³ = 0,17 moldm^-3

B. What if there would be a 0,2mol/l solution of NaOH instead of the 0,1mol/l solution?

NaOH

C_NaOH = 0,2 moldm^-3

V_NaOH = 0,034 dm³

N_NaOH = ?

n_NaOH = 0,2 moldm^-3 * 0,034 dm³ = 0,0068 mol

HCl

v_HCl = 0,02dm³

n_HCl = 0,0068 mol (n_NaOH = n_NaOH !)

c_HCl = 0,0068 mol / 0,02dm³ = 0,34 moldm^-3

To get to a neutral solution with these substances the concentration of HCl has to be 0,34 mol/l or we have to use it the double amount, 0,04 dm³.

THE TITRATION

There were no problems, when doing the titration with the help of the these calculations.