Question

= O ACIDS AND BASES Calculating the pH at equivalence of a titration A chemist titrates 60.0 mL of a 0.1536 M butanoic acid (HC,H,CO2) solution with 0.8565 M NaOH solution at 25 °C. Calculate the pH at equivalence. The pk, of butanoic acid is 4.82. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH = x ?

Answer 1

Calculating the bH at equivalence of a titration. 60 ml of 0.1536 M CHE COOH and 0.8565M NaOH and bka = 4.89 Giren.

Initial milli moles of NaCOOH- MX VOL) -0.1536 M X 60mL 9.216 m. mal

- C3H₃COONa + H₂O Reaction : CCOOH + NaOH at equivalence point; initial milli moles of Chly coolt = millimales of NaOH (initial)

9.216 m.mol = VNQOH soda X 0.8565M Vsoln (NaOH) total 9.2.16 ML = 10.76 mL 0.8565 Hence, Total volume of the soln, vt Vacia + Base at equivalence point, Vtotal Initial millimoles of Cathy Coop = finalm, mol of th cooNa = 9.216 manal molarity of C3H2COONa = m.mol -0.13M 70.76mL = 60mL + 10.76m2 = 70.76 m2 9.216 m.mol Vtotal

ICE Table initial(mol) g. 816 m.mol C₂ H₂COOH a NaOH C₂ H₂ cooNa + H₂O 9.216 m.mal Changelm.mel) -9.916 -9.216 + 9.216 Equilibrium 9.216 (m. mod) Hers only SILCOONa is present. Herrace How will depends on the hydrolysis of SHCOONa. C3H₃COONa + H₂O = CH₃COOH + NaOH intiallym) 0.13 change (m) equilibrium 0.13-4 Kw Ko Ka # -x x [CH3COOH) ROH] Kb Kw Ка (снасоо м 2 low kw Ka 2.31 (0.13 - x) as I will be very henee, 0.13-2 ~ 0.13

Kw Ka 22 δη 0.13 OY 22 0.13 X Kw ka take log both side Kw Ka ka loga? log 6.13 a log a = log(0.13) + log Kw - multiply by (-1) on the both side af log x) = -log (0.13) - log (1044)-(-log ka), a [logow])= 0 = 0.88 +14 -” ka x= [OH-] = [Naot] (DOH): 14.88-4.82 -Ioga= -loglovi]=bot s as kw=10.14 OY and OY g (pon) = 10.06 pOH = 5.03 as we know that bi thot 14 ÞOH = 14- ÞH or ÞH = 14-POH Therefore, pH = 14-5.03=8.97 þ Hi 8.97 Answer the pH at equivalence of a titration is/pH= 8.97

At equivalence point of reaction, the initial millimoles of the reactants and the final millimoles of the products are equal.

First, determine the volume of NaOH solution and then the volume of the total solution.

Then find the molarity of the salt$\large [C_3H_7COONa]$

Now, the pH of the solution depending on the hydrolysis of the $\large [C_3H_7COONa]$ ,

Using the formula $\large pOH=\frac{pK_b-\log [C_3H_7COONa]}{2}$

Found to be pOH =5.03

As we all know that pH + pOH = 14,

pH =14-pOH=14-5.03=8.97

Answer: the pH at the equivalence point is 8.97.