Question

1)Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)?

The K_a for HOCl is 3.0 x 10^-8 M.

2)How many mL of NaOH are added to reach the equivalence point

3) What is the pH after 2.40 mL of NaOH are added?

Answer 1

1.Initial pH of solution or at 0.0 ml Base addition HOCI acid is a weak acid, it dissociate partially into acetate and H+ in aqueous media as shown below HOCI (aq) <-----> OC1- (aq) + H+ (aq) Initial 0.004 Change - X Equilibrium 0.004 0 + - x) Assume x is small then 0.004 - x will be = 0.004 M 0.004 Given Ka= 3.00E-08 Weak acid dissociation constant Ka=(OC1-)(H+)/(HOCI) 3.00E-08 x*x+(0.004 3.00E-08 = x*x+ 0.004 x^2 = 3.00E-08 sox^2 1.20E-10 0.00001 but from ICE table x = [H+]; So [H+] = 0.0000110 but pH = -log [H+] = -log[ 0.00001095 ] = 4.96041 so pH of solution = 4.96 M 2. After addition of 2.4 ml base Given Molarity of HOCI = 0.400 M Volume of HOCI 10.00 ml = 0.010 L We know, Moles = Molarity (M) x Volume (V in L) Moles of HOCI= 0.400 Mx 0.0100 L = 0.004 mol 0.0024 L Volume of NaOH = 2.4 ml or Molarity of NaOH = 0.500 M Moles of NaOH = 0.500 mol L-1* 0.0024 L = 0.00120 mol Ka= 3.00E-08;pKa = log (ka) = -log [ 3.00E-08 ] = 7.5229 When nu: We know both HOCI acid and aceate act as a buffer system. The ICE table is shown below. HOCI (aq) + OH- (aq) <----->OCI- (aq) + H20 (1) Initial 0.0040 0.0012 0.000 Change -0.0012 -0.0012 +0.00120 Equilibrium 0.00280 0 0.00120 + Henderson equation for buffer solution is pH=pKa +log [Base Acid] Here Base is OCl- and acid is HOCI so pH =pKa + log[Base/acid] =pKa + log[OC1- / HOCI] = 7.523 log [ 0.00120 = 0.00280] = 7.523 log [ 0.4286 ] 7.523 + -0.3680 7.15 + +

pH at equivalent point Given Molarity of HOCI = 0.400 M Volume of HOCI 10.00 ml = 0.010 L We know, Moles = Molarity (M) x Volume (V in L) Moles of HOCI = 0.400 MX 0.0100 L = 0.004 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) 0.0080 L Moles of NaOH needed to reach equivalent = 0.00400 mol Molarity of NaOH given = 0.500 M Volume of NaOH needed to reach equilibrium = Moles of Base / Molarity of Base = 0.0040 mol = 0.500 mol L-1 = 0.0080 L Total volume of solution at equivalent point = Volume of Acid + Volume of Base = 0.01000 L + = 0.018 L ICE table for equivalent point can shown as below HOCI (aq) + OH- (aq) <-----> OC1- (aq) + H2O (1) Initial 0.00400 0 .00400 0.0000 Change -0.00400 -0.00400 + 0.00400 Equilibrium 0.0 0 0.00400 Above ICE table shows only presence of acetate, and moles of acetate = 0.0040 mol Concnetration of Acetate OCl- = Moles of OC1- / total volume = 0.00400 mol = 0.0100 L = 0.400 M Now the pH of OCi- can be calculated as shown below OH- (aq) 0.0 Initial Change Equilibrium OC1-(aq) + H2O (aq) <-----> HOCI(aq) + 0.4000 0.0 (-x) (+x) (0.400-x) (+x) Given Ka of HOCI= 3.000E-08; Kw= 1.00E-14 Then Kb = Kw/Ka= 1.0 x 10^-14 = 3.00E-08 = 3.33E-07 Acetate is a weak base and equilibrium constant Kb can be expressed as below Kb = (HOCI X OH-) =( C3H302-) 3.33E-07 = (x) (x) (0.400-X asume x is small then so 3.33E-07 x^2 = 0.400 (0.400-x) will be 0.4000 x^2 = 3.33E-07 * 0.4000 sox^2 1.333E-07 3.651E-04 but from ICE table x = [OH-] So [OH-] = 3.651E-04 but pOH = -log [OH-] = -log [ 3.651E-04 ] = 3.4375 we Know pH + pOH = 14 or pH = 14- pOH = 14 - 3.4375 = 10.5625 pH of solution at equivalent point = 10.56