hem 1120 Name: Homework 6 (30 pts total): Due Tuesday, July 26, 2016 A chemist i

Question

hem 1120 Name: Homework 6 (30 pts total): Due Tuesday, July 26, 2016 A chemist is working to identify a unknown weak acid. dissolves 1.00 g of the weak acid in enough water to make a total volume of the mL and titrates the solution with 0.2498 NaoH. general shape of titration curve looks like this (figure is intentionally vague) The pH of the original solution (before adding any NaoH) is 2.257. The solution requires 4444 mL ofNaoH to each the equivalence point. First, find the number of moles of acid in 1.00 g, using the volume of base added at the equivalence point. i pt moles 0.2 b) So what is the molar mass? pt) molar mass: c) Now that you know the concentration of the acid in the original solution, use the starting pH to find the Ka and pKa of the acid. (2 pts) 2.757 K. d) What is the pH ofthe titration solution at the equivalence point? (Hint at the equivalence point, all HA has been converted to A. So solve the equilibrium problem for the conjugate base. Don't forget to adjust the concentration to account for the added volume of Naoh). G pts)

Explanation / Answer

c.

note that

pKa = -log(Ka)

Apply

Ka = [H+][A-]/[HA]

get all the equilibrium concentrations

pH = 2.257

[H+] = 10^-pH = 10^-2.257 = 0.005533 M

Since, [H+] = [A-] due to sotichiometry

then

[A-] = 0.005533

and

[HA] = M-x

M = mol of acid / V = 0.011 / 0.050 = 0.22 M

x = 0.005533

[HA] = M-x = 0.22-0.005533 = 0.214467

then

Ka = (0.005533)(0.005533)/0.214467 = 0.00014274498 = 1.427*10^-4

pKa = -log(1.427*10^-4) = 3.8845

d.

in equivalence point

we have a very peculiar point

pH = -log(H+)

A- + H2O <-> HA + OH-

therefore expect basic medium

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka = (10^-14)/(1.427*10^-4) = 7.01*10^-11

[HA] = OH- = x

Vt = 50 + 44.44 = 94.44 ml (account for the acid +base volumes)

[A-] = mol/Vt = 0.011/0.09444 = 0.116476

substitute all

Kb = [HA][OH-]/[A-]

7.01*10^-11 = x*x / (0.116476)

x = sqrt((7.01*10^-11)*0.116476)

X = 0.00000285744

[OH-] = 0.00000285744

pOH = -log(OH) = -log(0.00000285744 = 5.5440

pH = 14-pOH = 14-5.5440

pH = 8.456

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