MODULE ANSWER

MODUL ANSWER

Preparation of a solution by dilution (pg 128)

1(a) Standard solution is a solution in which its concentration is accurately known.

(b)        -molar mass of NaOH = 23+16+1 = 40 g mol^-1

              - mol NaOH              = 500 × 1.0 /1000 = 0.5 mol

              - Mass of NaOH        = 0.5 mol × 40 g mol^-1= 20.0 g

Preparation of 500 cm³  1.0 mol dm^-3 sodium hydroxide

-          20.0 g                          weighing bottle

-          20.0 g                          distilled water

-          Volumetric flask

-          Rinse           distilled water   volumetric flask

-          Distilled water         volumetric flask                                calibration mark

-          Volumetric flask     stopper                                inverted

Calculate the volume of 1 mol dm^-3 sodium hydroxide used:

-          M₂  × V₂      = 0.1 × 250 = 25 cm³

M1                   1

Preparation of 250 cm³ 1.0 mol dm^-3 sodium hydroxide

-          25 cm³

-          25 cm³         volumetric flask

-          Distilled water         volumetric flask                                calibration mark

-          Volumetric flask     stopper                                inverted

1(a) (i) weak acid       : An acid that partially ionizes in water to produce low concentration of hydrogen  ion, H⁺.

Strong acid                  : An acid that completely ionizes in water to produce high concentration of hydrogen ion, H⁺.

(ii)

-          higher

-          stong acid                  completely                         higher

-          weak acid                  partially

-          CH3COO^-  +   H⁺

(iii)

-          higher                         lower

-          lower                          higher

b)

  -      molecules                  ionise                    molecules           molecules           neutral

   Hydrogen

-          ionises partially       ethanoate           hydrogen

2a(i) solution P

     (ii) solution U

b(i)         Q

(ii)         R

(iii)        T

(iv)       P

(v)        U

(vi)       S

c(i) P/Q/R and T/U

  (ii) P/Q

3.       80 g dm^-3

4.       0.5 mol dm^-3

5.       2 mol dm^-3

6.       10 g

7.       0.4 mol dm^-3

8.       50 cm³

9.       a)

Number of mol of sulphuric acid  = 50 × 1       1000  = 0.05 mol H2SO4 → 2H+ + SO42- From the equation, 1 mol of H2SO4: 2 mol of H+ 0.05 mol of H2SO4 : 0.1 mol of H+	Number of mol of hydrocloric acid  = 50 × 1       1000  = 0.05 mol HCl → H+ + Cl- From the equation, 1 mol of HCl: 1 mol of H+ 0.05 mol of HCl : 0.05 mol of H+
The number of H+  in 50 cm3 of 1 mol dm-3 of sulphuric acid is twice of the number of in 50 cm3 of 1 mol dm-3 of hydrochloric acid.
Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid. 1 mol of sulphuric acid ionises to 2 mol of H+  whereas 2 mol of hydrochloric acid ionizes to 1 mol of H+ . The number of H+  in the same volume and same concentration is doubled in sulphuric acid compared to hydrochloric acid.

b) 100 cm³