Question

Acer C Example (Tutorial) point is point, the Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 ml. tiurant in the titration of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH K 1.75 x 10 Ac] and ses as

Answer 1

Given Molarity of HC2H302 = 0.100 M Volume of HC2H302 = 50.00 ml = 0.050 L We know, Moles = Molarity (M) x Volume (V in L) Moles of HC2H302 = 0.100 MX 0.0500 L = 0.005 mol Molarity of NaOH = 0.100 M 1.Initial pH of solution or at 0.0 ml Base addition Acetic acid is a weak acid, it dissociate partially into acetate and H+ in aqueous media as shown below HC2H302 (aq) <-----> C2H302 - (aq) + H+ (aq) Initial 0.005 0 Change Equilibrium 0.005 - X + x Assume x is small then 0.005 - x will be = 0.005 M x^2 Given Ka= 1.75E-05 Weak acid dissociation constant Ka=(C2H302 -)(H+)/(HC2H302) 1.75E-05 = x*x+( 0.005 - x) 1.75E-05 x*x= 0.005 1.75E-05 * 0.005 SO X^2 = 8.75E-08 0.00030 but from ICE table x = [H+]; So [H+] = 0.0002958 M but pH = -log [H+] = -log[ 0.00029580 ] = 3.529 SO pH of solution= 3.53 2. After addition of 10.0 ml base Given Molarity of HC2H302 = 0.100 M Volume of HC2H302 50.00 ml = 0.050 L We know. Moles = Molarity (M) x Volume (V in L) Moles of HC2H302 = 0.100 Mx 0 .0500 L = 0.005 mol 0.0100 L Volume of NaOH = Molarity of NaOH = Moles of NaOH = 10.0 ml or 0.100 M 0.100 mol L-1* 0.0100 L = 0.00100 mol Ka= 1.75E-05: pKa = log (ka) = -log [ 1.75E-05 ] = 4.757 When OH- is added to acetic acid, it reacts with acetic acid, and produce aceate. We know both acetic acid and aceate act as a buffer system. The ICE table is shown below. HC2H302 (aq) + OH- (aq) <-----> C2H302- (aq) + H20 (1) Initial 0.0050 0.0010 0.000 Change -0.001 -0.0010 +0.00100 Equilibrium 0.00400 0.00100 Henderson equation for buffer solution is pH=pKa +log [Base Acid] Here Base is C2H302- and acid is HC2H302 so pH =pKa + log[Base/acid] =pKa + log[C2H302- / HC2H302] = 4.757 + log [ 0.00100 0.00400 ] 4.757 + log [ 0.2500 ] = 4.757 -0.6021 4.15 3. pH at half equivalent point or 25 ml Given Molarity of HC2H302 = 0.100 M Volume of HC2H302 = 50.00 ml = 0.050 L We know, Moles = Molarity (M)x Volume (V in L) Moles of HC2H302 = 0.100 Mx 0.0500 L = 0.0050 mol 0.0250 L Volume of NaOH = 25.0 ml or Molarity of NaOH = 0.100 M Moles of NaOH = 0.100 mol L-1* 0.0250 L = 0.00250 mol Ka= 1.75E-05;pka = log (ka) = -log[ 1.75E-05 1 = 4.757 When OH- is added to acetic acid, it reacts with acetic acid, and produce aceate. We know both acetic acid and aceate act asa a buffer system. The ICE table is shown below. HC2H302 (aq) + OH- (aq) <-----> C2H302- (aq) + H20 (1) Initial 0.00500 0 .00250 0.000 Change -0.0025 +0.00250 Equilibrium 0.00250 0.00250 -0.0025 Henderson equation for buffer solution is pH=pKa + log [Base/Acid] Here Base is C2H302- and acid is HC2H302 so pH =pKa + log[Base/acid] =pKa + log[C2H302-/ HC2H302] 4.757 + log 0.00250 0.00250 1 = 4.757 + log [ 1.0000] = 4.757 + 0.0000 = 4.76 so pH at half equivalent point = 4.76

4. pH at equivalent point Given Molarity of HC2H302 = 0.100 M Volume of HC2H302 = 50.00 ml = 0.050 L We know, Moles = Molarity (M) x Volume (V in L) Moles of HC2H302 = 0.100 MX 0.0500 L = 0.005 mol At equivalent point initial moles of weak acid will be completely nutralized and same amount of conjugate base will be formed and Moles of base need to reach equivalent point = Moles of Acid (since it is a mono protic acid) 0.0500 L Molarity of NaOH given = 0.100 M Volume of NaOH = 50.00 ml = 0.050 L Moles of NaOH = 0.100 MX 0.0500 L = 0.005 mol Total volume of solution at equivalent point = Volume of Acid + Volume of Base = 0.05000 L + 0.1000 L ICE table for equivalent point can shown as below HC2H302 (aq) + OH- (aq) <-----> C2H302- (aq) + H20 (1) Initial 0.00500 0.00500 0.0000 Change -0.00500 -0.00500 + 0.00500 Equilibrium 0.0 0.00500 Above ICE table shows only presence of acetate, and moles of acetate = 0.0050 mol Concnetration of Acetate C2H302- = Moles of C2H302-/ total volume = 0.00500 mol + 0.1000 L 0.050 M Now the pH of C2H302- can be calculated as shown below 0 OH- (aq) 0.0 Initial Change Equilibrium C2H302 - (aq) + H2O (aq) <-----> HC2H302(aq) + 0.0500 0.0 (+x) (0.050-x) (+x) Given Ka of HC2H302 = 1.750E-05; Kw= 1.00E-14 Then Kb = Kw/Ka= 1.0 x 10^-14 1.75E-05 = 5.71E-10 Acetate is a weak base and equilibrium constant Kb can be expressed as below Kb = (HC2H302 x OH-)+(C3H302-) 5.71E-10 = (0.050-x) SO 5.71E-10 x^2 = 0.050 5.71E-10 * 0.0500 so x^2 2.857E-11 5.345E-06 but from ICE table x = [OH-] so [OH-] = 5.345E-06 but pOH = -log [OH-] = -log [ 5.345E-06 ] = 5.272 asume x is small then (0.050-x) will be 0.0500 we Know pH + pOH = 14 or pH= 14- pOH = 14 - 5.272 = 8.72797 pH of solution at equivalent point = 8.73 5. pH at after equivalent point at 60 ml Molarity of HC2H302 = 0.1000 M Molarity of NaOH = 0.1000 M Volume of HC2H302= 50.000 ml = 0.0500 L Vol of NaOH = 60.00 ml = 0.0600 L Moles of HC2H302 = 0.100 X 0.05000 Moles of NaOH = 0.1000 X 0 .0600 L = 0.0050 mol = 0.0060 mol Note: Moles = Molarity x volume Total volume of solution at equivalent point = Volume of Acid + Volume of Base = 0.05000 L + 0.0600 L = 0.11000 L HC2H302 (aq) + OH- (aq) <-----> C2H302- (aq) +H20 (1) Initial 0.00500 0.0060 0.000 Change -0.00500 -0.00500 +0.00500 Equilibrium 0.0 0.0010 0.00500 After addition of excess OH-, ICE table shows presence of C2H302- and OH-; Since C2H302- is a weak base, so we can neglect C2H302- while calculating pH of solution So Conc of OH-= Moles of OH-/Total volume = 0.0010 mol + 0.11000 L = 0.0091 mol L-1 So [OH-] = 0.0091 M We know pOH = -log [OH-] = -log[ 0.0091 ] = 2.0414 we Know pH + pOH = 14 or pH = 14- pOH = 14 - 2.0414 = 11.96 so pH after addition of 60.00 ml NaOH = 12.0