Exercise 16.65 Consider the following curve (Figure 1) for the titration of a we

Question

Exercise 16.65 Consider the following curve (Figure 1) for the titration of a weak monoprotic acid with a strong base and answer each of the following questions Part A What is the pH at the equivalence point? Express your answer as whole number. Part B What is the volume of added base at the equivalence point? Express your answer using one significant figure. Submit My Answers Give Lie Part C At what volume of added base is the pH calculated by working an equillibrium problem based on the ii Express your answer using one significant figure. Part D At what volume of added base does pH = pKa? Express your answer using two significant figures.

Explanation / Answer

A)

From the graph

equivalence point is chieved approx at V = 30 mL

the pH is given approx at 8.5-9

pH = 9

B)

volume of acid is approx V = 30 mL of acid

C)

equilbirium problem is worked when V = 0, for acid

D)

part in which

pKa = pH

is given as

pH = pKa + log(a-/HA)

the buffer equation

so

pH = pKA only when A- = HA

which is at the HALF equivalence point

so

V = 30/2 = 15 mL approx

Question 2.

B)

volume of lood = 5 L

find MASS of HCl requied for neutralization

HCO3- + HCl = H2CO3 + Cl-

then

mol of HCO3- = MV = 0.024 * 5 = 0.12 mol of HCO3-

then we need

0.12 mol of HCl

mass = mol*MW = 0.12*36.5 = 4.38 g of HCl required

C)

volume of B for

pH = 7.8

pH = pKa + log(HCO3-/H2CO3)

7.8 = 6.1 + log(HCO3-/H2CO3)

(HCO3-/H2CO3) = ratio = 10^(7.8-6.1) = 50.11

as we add NaOH:

mol of NAOH

mol of HCO3- = MV = 0.024*5 = 0.12

mol of H2CO3 = MV = 0.0012*5 = 0.006

50.11 = (0.12 + x)/(0.006-x)

50.11*0.006 - 50.11x = 0.12 + x

(50.11+1)x = 50.11*0.006-0.12

x = (50.11*0.006-0.12 ) / ((50.11+1))

x = 0.0035 g

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