Question

Please Answer the Following Questions (#1-4)

#1 You are proved a buffer whose acid pKa is 6.17 and solutions of 1M Hal and 1M NaOH. The initial pH of this buffer is 5.24. Calculate how many milliliters of Hal or NaOH are required to alter the buffer's pH to 6.65 and assume that [HA] and [A-] concentration are in mol units.

#2

You are given a 25mL sample of weak acid (HA) of concentration 0.153M, with a Ka of 5.6*10-7 Titrate this with a 0.18M NaOH titrant. Calculate the pH values for the following values of titrant:

a. Volume at 0.00mL

b. Volume at 9.5mL

c. Volume at 17.25mL

d. Volume at equivalence point

e. Volume at 30mL

#3 You are asked to make a buffer using a weak acid (HA, FM=119.96 g/mol; Ka=1.63*10^-6) and its conjugate base (A-, FM=141.94 g/mol). You place 4.167g of HA and 6.555 g A- into a flask and dilute to 250mL Calculae the pH of this buffer?

#4 What is the pH of a 1.18 M solution of the intermediate form (e.g., HA- or BH+) of 2-aminophenol?

Answer 1

Question 2.

[HA] = 0.153 M, [NaOH] = 0.18 M

pKa = -Log(Ka) = -Log(5.6*10^-7) = 6.25

(a) 0 mL of base is added

i.e. pH = 1/2 (pKa - Log[HA]) = 1/2 (6.25 - Log0.153) = 3.53

(b) 9.5 mL of base is added

ii.e. pH = pKa + Log{[NaA]/[HA]} = 6.25 + Log(9.5*0.18/(25*0.153 - 9.5*0.18)} = 6.16

(c) 17.25 mL of base is added

ii.e. pH = pKa + Log{[NaA]/[HA]} = 6.25 + Log(17.25*0.18/(25*0.153 - 17.25*0.18)} = 6.88

(d) Volume at equivalence point (V = 25*0.153/0.18 = 21.25 mL)

[NaA] = 25*0.153 mmo/(25+21.25) mL = 0.0827

pH = 7 + 1/2 {pKa + Log[NaA]} = 7 + 1/2 {6.25 + Log(0.0827)} = 9.58

(e) [NaOH] = [OH^-] = {0.18*30 - (0.153*25)}/(30+25) = 0.0286 M

Now, pH = 14 - {-Log[OH^-]} = 14 - {-Log(0.0286)} = 12.46