Answer:
Step-by-step explanation:
1) Acetic acid
Concentration is given as 0.103 M
The average pH of this solution = 2.96
we know that pH = - log [H+] therefore [H+] = 10-pH
[H+] = 10-2.96
= 1.1 x 10-3 M = 0.0011 M
Consider the equilibrium
CH3COOH ⇄CH3COO- + H+
Initial 0.103 0 0
Change -x +x +x
equlibrium 0.103 -x x x
Ka = x2 / 0.103 - x
Here the initial concentration of CH3COOH = 0.103 M
the equilibrium concentration of H+ = x = 0.0011 M
Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M
Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5
2) Acetic acid + NaOH
pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5
Volume and conc of acetic acid = 10 mL of 0.103 M
= 10 mL x 0.103 mmol / mL
= 1.03 mmol
Volume and conc of NaOH added = 4 mL of 0.0992 M
= 4 x 0.0992 mmol
= 0.397 mmol
Consider the equation
CH3COOH + NaOH -----------> CH3COONa + H2O
Initial 1.03 0.397 0
Final 0.633 0 0.397
0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.
Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397
therefore [salt] / [acid] = 0.397 / 0.633 = 0.627
By Hendersen equation pH = pKa + log[salt / acid]
pH = pKa + log 0.627 = pKa - 0.203
or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]
= 4.683
Ka = 10-4.683 = 2.07 x 10-5
3) Phosphate salts:
(i) mass of NaH2PO4 taken = 0.613 g
molar mass of NaH2PO4 = 120
therefore moles = 0.613 / 120 = 0.0051 mole
= 5.1 mmol
The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL
= 0.17 M
consider the equilibrium
H2PO4-⇄ HPO42- + H+
Initial 0.17 0 0
Change -x +x + x
equilibrium 0.17-x x x
Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]
neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8
x = 1.03 x 10-4
Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99
(ii) Mass of Na2HPO4.7H2O =0.601 g
therefore no of moles = 0.601 / 268.07 = 0.00224 mole
= 2.24 mmol
The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml
= 0.075 M
consider the equilibrium
HPO42- ⇄ PO43- + H+
Initial 0.075 0 0
Change -x +x + x
equilibrium 0.075-x x x
Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]
neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13
x = 1.9 x 10-7
Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7
(iii) Mass of Na3PO4.12H2O taken = 0.208 g
moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles
= 0.55 mmol
Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL
= 0.055 M
Consider the equilibrium reaction
PO43- + H2O ⇄ HPO42- + OH-
initial 0.055 0 0
Change -x +x +x
equilibrium 0.055-x x x
Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]
x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025
That is the conce of OH- ion = 0.025M
Therefore pH = 14 - pOH = 14 - 1.6 =12.4