Question

50 ml of 0.100M solution of a weak acid HB titrated with NaOH. Calculate ph at the start, after 10.0 ml, 50.0 ml, and 60.0 ml.

Ka=1.0x10^-5

[NaOH]=0.1M

Answer 1

0.050 L Given Molarity of HA = 0.100 M Volume of HA = 50.00 ml = We know, Moles = Molarity (M) x Volume (V in L) Moles of HA = 0.100 MX Molarity of NaOH = 0.100 M 0.0500 L = 0.005 mol 1.Initial pH of solution or at 0.0 ml Base addition HA a weak acid, it dissociate partially into A- and H+ in aqueous media as shown below HA(aq) <-----> A- (aq) + H+ (aq) Initial 0.005 0 Change + x Equilibrium 0.005 - X - x) Assume x is small then 0.005 - x will be = 0.005 M x^2 * 0.005 Given Ka= 1.00E-05 Weak acid dissociation constant Ka=(A-)(H+)/(HC2H302) 1.00E-05 x x=( 0.005 1.00E-05 x*x= 0.005 1.00E-05 sox^2 5.00E-08 0.00022 but from ICE table x = [H+]; So [H+] = 0.0002236 but pH = -log [H+] = -log[ 0.00022361 ] = 3.65051 so pH of solution= 3.65 M 2. After addition of 10.0 ml base Given Molarity of HA = 0.100 M Volume of HA 50.00 ml = 0.050 L We know. Moles = Molarity (M) x Volume (V in L) Moles of HA= 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= 1.00E-05 : pKa = log (ka) = -log[ 1.00E-05 ) = 5.000 We know both HAand A- act as a buffer system. The ICE table is shown below. HA(aq) + OH- (aq) <-----> A- (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 A- and acid is HC2H302 so pH =pKa + log[Base/acid] =pKa + log[A-/HC2H302] 5.000 log [ 0.00100 0.00400 ] = 5.000 log [ 0.2500 ] 5.000 -0.6021 4.40 + +

4. pH at equivalent point or 50 ml Given Molarity of HA = 0.100 M Volume of HA 50.00 ml = 0.050 L We know, Moles = Molarity (M) x Volume (V in L) Moles of HA = 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 Moles of NaOH needed to reach equivalent = 0.00500 mol Molarity of NaOH given = 0.100 M Volume of NaOH needed to reach equilibrium = Moles of Base / Molarity of Base = 0.0050 mol = 0.100 mol L-1 = 0.0500 L Total volume of solution at equivalent point = Volume of Acid + Volume of Base = 0.05000 L + = 0.100 L ICE table for equivalent point can shown as below HA(aq) + OH- (aq) <-----> A- (aq) + H20 (1) Initial 0.00500 0.00500 0.0000 Change -0.00500 -0.00500 + 0.00500 Equilibrium 0. 00 0.00500 Above ICE table shows only presence of A-, and moles of A- = 0.0050 mol Concnetration of A-A- = Moles of A- / total volume = 0.00500 mol = 0.1000 L = 0.050 M Now the pH of A- can be calculated as shown below + OH- (aq) A-(aq) + H2O (aq) <-----> 0.0500 HA (aq) 0.0 (+x) Initial Change Equilibrium 0.0 (0.0500-x) Given Ka of HA= 1.00E-05 ; Kw= 1.000E-14 Then Kb = Kw/Ka= 1.0 x 10^-14 = 1.00E-05 = 1.00E-09 A- is a weak base and equilibrium constant Kb can be expressed as below Kb = (HAX OH-) =( C3H302-) 1.00E-09 (x) (x) (0.0500-x) 1.00E-09 x^2 = 0.050 1.00E-09 * 0.0500 sox^2 5.000E-11 7.071E-06 but from ICE table x = [OH-] So [OH-] = 7.071E-06 but pOH = -log [OH-] = -log [ 7.071E-06 ] = 5.1505 asume x is small then 10.0500-x) will be 0.0500 x^2 we Know pH + pOH = 14 or pH = 14-pOH = 14 - 5.1505 = 8.84949 pH of solution at equivalent point = 8.85 5. pH at after equivalent point Molarity of HA= 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 HA= 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 -- HA(aq) + OH- (aq) <-----> A- (aq) + H2O (1) Initial 0.00500 0.0060 0.000 Change -0.00500 -0.00500 +0.00500 Equilibrium 0.0010 0.00500 After addition of excess OH-, ICE table shows presence of A- and OH-; Since A- is a weak base, so we can neglect A- 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 = 11.96