Chemistry Paper 3B May/June 2015

Question 1B

F is a solution of NaOH containing 0.0500 mol of the alkali in 500 cm3 of solution.         
            G is a solution of a monobasic acid, HY.
                       
(a)        Put solution G into the burette. 

Titrate 20.0cm3 or 25.0 cm3 portions of F   against G using phenolphthalein as indicator. 

Repeat the exercise to obtain consistent titres.

Tabulate your burette readings and calculate the average volume of G used.

                        The equation of reaction is:

                NaOH(aq)  +    HY(aq)    chemimg04       NaY(aq)  +  H2O(l)

            (b)        From your results and the information provided, calculate the:

(i)         concerntration of F in moldm-3;
(ii)        concerntration of G in moldm-3;
(iii)       molar mass of the acid, HY if the concentration of acid in solution G is                                          3.95 gdm-3
(iv)       relative atomic mass of Y;
(v)        percentage by mass of Y in HY. [H = 1.00]                                                                               [21 marks]

Observation

 

In the part (b), majority of the candidates showed good skills in the calculation of the concentration of F and G in moldm-3 respectively, but failed to give the correct substitution
of 3.95gdm-3 when they were calculating the molar mass of the acid, HY.

Expected answers include:

1(a)      Two consistent / concordant titres Averaging

(b)

(i)         Conc. of Fin moldm-3
Conc. of F (NaOH)     =       0.050 x 1000 OR 0.050 × 2
500      
=        0.100moldm-3

(ii)        Conc. of G in moldm-3
Conc. of G(HY)  =   CGVG      =                    
CFVF                                        
CG   =  CFVF         =  0.100x 20.0/25.0            
VG                         VG

                                                = amol dm-3

Alternative Method
Amount of base in 20/25 cm3 B = 0.100  x  V           
1000
= 0.001 VBmol
From balance equation of reaction
1mol of B ≡ 1mol of A
 0.0001VB of B ≡ 0.0001VBmol of A
VAcm3   ≡ 0.0001VB × 100mol
VA
= 0.1VB  mol
VA          
Conc. of A           = 0.1VB  moldm-3
                                    VA                              

(iii)       Molar mass of the acid G(HY)
CG =
MG(HY) =    =  3.95
CG           a
                = bgmol-1                                                                            

(iv)       Relative atomic mass of Y in HY
HY = b         
1 + Y = b     
  Y = b – 1
= X say    

(v)        Percentage by mass of Y in HY
% of Y in HY = X × 100   
b

                        =  z