Question

Buffers and pH Changes Tube pH Before Addition pH After Addition PH Change 4,5 pH Water + Acid Water + Base Buffer + Acid Buffer + Base S.SoH s.sph Tors of SpH 2 7 8 H PH Hydrolysis of lonic Compounds 2 NaCl NHACI FeCl3 pH 4-5 5 pH 3-4 PH 1-2 PA Na2CO3 Prediction pH 10-11 pH Measurement Il Titration of Sodium Carbonate Formula of Compound: Nazlog pH 1 매 니 0.432 Mass of Compound: 0.004 Vol. pH PH 1 ml 10.80 2 ml 3 mL 4 ml 5 mL 10-31 Moles of Na2CO3: kritial PH: 11.05 Vol pH Vol Vol pH 11 mL 9.66 21 mL 7.15 31 mL6.05 12 mL 9.58 22 mL 6.96 32 mL 5.96 13 mL 9.48 23 ml 6.81 33 mLS.86 14 mL 9.31 24 mL 6.69 34 ml 5.76 15 ml 9.27 25 ml 6. S7 35 mL 5.65 16 mL 9.14 26 mL 6.47 17 ml $.94 27 ml 6.38 37 ml 5.30 18 ml 8.70 28 mL 6.21 38 ml 19 mL 8.24 29 mL 6. 22 20 ml 7.40 30 ml 6.12 40 mL 2.58 Vol PH 41 ml 2.35 42 mL 12:18 43 mL 2.08 44 mL 2,02 45 mL 1.93 46 mL 7.87 47 mL 7.82 48 mL/.78 49 mL 7.74 50 mL 1.71 6 mL 36 mL 5.49 7 ml 8 ml 10.09 10.07 9.93 9.84 q76 14.92 9 ml 39 mL 3.29 10 mL

Answer 1

Answer to 3A

Hydrolysis of Na2CO3, $CO_3^{2-}$ will undergo hydrolysis

CO3- (aq) + H2O(l) = HCO3(aq) + OH-(aq)

NaCl won't hydrolysis,as Na+ and Cl- both are inert to hydrolysis, they won't react with water so the pH given in that box is not correct. In this case pH must be neutral which is 7 at 25$^0C$

Hydrolysis of NH4Cl

will undergo hydrolysis as

NH

$NH_4^+(aq)+H_2O \rightleftharpoons NH4OH (aq)+H^+(aq)$

Hydrolysis of feCl3, Fe3+ will hydrolyses as

$Fe^{+3}(aq)+H_2O \rightleftharpoons Fe(OH)^{+2} (aq)+H^+(aq)$

ANSWER TO 3(B)

First neutralisation reaction

$Na_2CO_3 (aq)+HCl(aq) \rightarrow NaHCO_3(aq)+NaCl(aq)$

In ionic form, $CO_3^{2-} (aq)+H^+(aq) \rightarrow HCO_3^-(aq)$

Second neutralisation reaction

$NaHCO_3 (aq)+HCl(aq) \rightarrow H_2CO_3(aq)+NaCl(aq)$

In ionic form, $HCO_3^{-} (aq)+H^+(aq) \rightarrow H_2CO_3(aq)$

Decomposition of H2CO3

$H_2CO_3(aq) \rightleftharpoons H_2O(l) + CO_2(g)$

Answer to 3C and 3D

Titration of Sodium Carbonate 12 . A [Only CO2 10 This step change indicates 1st equivalence point Ci.e. ist neutralisation) 'B 6 Call co² HCo) c [HCO3 → H₂CO N 2nd Equivalence pant 0 0 10 20 30 40 50 60 HCI (mL)

Titration of Sodium Carbonate D' 12 [002] : [H0] 10 8 E'CH80] CH, CO] 6 Buffer #1 4 Buffer #2 2 0 0 10 20 30 40 50 60 HCI (mL) the * Buffer is solution that contains both weak acid and its conjugate base Cor weak base and its Conjugate acid). [ Mixture of CONJUGATE ACID-BASE PAIR CCABP) is BUFFER] So, in Case of above Neutralisation Reactions. CABP 2- H. HCO * at mid point of this neutralisation, [2002]. [hco,]. The region which contains both Co and HCO % buffer marked as Buffer #1. CABP HCO + M H₂ CO₂ * at mid point of this neutralisation, [HCO3] = [H₂CO3]. The H₂ CO₂ is buffer marked as Buffer #2. region which contains both and HCO

Answer to 3E

From the region indicated as buffer due to carbonate and bicarbonate ion. . pH = pKa of acid (here HCO3-) when concentration of both the carbonate and bicarbonate ion are equal and that is achieved when almost half of the carbonate is converted to HCO3-, so we can say pH at about midpoint of the first buffer region (marked in figure) gives the pKa of HCO3-.

Similarly, to calculate pka of H2CO3,, we can make use of buffer region created due to HCo3- and H2CO3. (as marked in figure), so pKa of H2CO3 = pH at about midpoint of this buffer region.

Titration of Sodium Carbonate 12 so, at mid-point, here lies, pH which & 9.76 from table) 8 PH =9.76 at vol. = 10 ml. 6 [HC05) [002] ... This pH = pka of HCO. 9.76 4 2 0 10 20 40 50 60 30 HCI (mL) Mid-point Titration of Sodium Carbonate 12 10 8 pH at 30 ml 6-12 Cfrom table) Thus, pka of H₂CO = 16.12) 6 4 2 0 10 20 30 50 60 HCL) midpoint and neutralisation

Answer to 3F

Thus,

pka of H2CO3 = 6.12, and $Ka = 10^{-6.12}=7.6 \times 10^{-7}$

pka of HCo3 - = 9.76 and $Ka = 10^{-9.76}=1.74 \times 10^{-10}$