Hi Beautiful-Mind! Please answer all parts: You are preparing to titrate a 30.0
ID: 902732 • Letter: H
Question
Hi Beautiful-Mind!
Please answer all parts:
You are preparing to titrate a 30.0 mL solution of 0.100 M 3-nitrophenol (HA) with 0.1000 M NaOH.
Ka = 4.3 x 10-9
a) Write the balanced titration equation
b) What is the initial pH of 3-nitrophenol solution
c) Calculate the volume at the equivalence point.
d) Calculate the pH at the equivalence point.
e) Select an appropriate acid/base indicator to find the end point for this titration. Predict the color change that will be observed at the end point.
Thanks in advance!
Explanation / Answer
V = 30 ml
M = 0.1 M HA
M = 0.1 M of NaOH
Ka = 4.3*10^-9
a) Write the balanced titration equation
HA + NaOH <--> NaA + H2O
b) What is the initial pH of 3-nitrophenol solution
initial pH
Ka = [H+][A-]/[HA]
4.3*10^-9 = x*x / (0.1-x)
solve for x
x = 2.07*10^-5
[H+] = 2.07*10^-5
pH = -log(H+) = -log(2.07*10^-5) = 4.68
c) Calculate the volume at the equivalence point.
equivalence point is when 1 mol of base : 1 mol of acid
M1*V1 = M2*V2
30*0.1 = 0.1*V
V = 30 ml of base are needed
Total Volume = V1+V2 = 30 + 30 = 60 ml
d) Calculate the pH at the equivalence point.
In the equivalnce
we have 0.1*30ml = 3 mmol of A- conjugate which willl hydrolyze
M = mol/V = 3 mmol/60 = 0.05 M
A- + H2O <--> HA + OH-
Kb = [HA][OH-]/[A-]
Kb = Kw/Ka = (10^-14)/(4.3 x 10-9 ) = 2.33*10^-6
in equilibrium:
[HA] = [OH-]= x
[A-] = 0.05 - x
Kb = [HA][OH-]/[A-]
2.33*10^-6 = (x*x)/(0.05-x)
solve for x
x = 3.4*10^-4
[OH-] = x = 3.4*10^-4
pOH = -log(OH) = -log(3.4*10^-4) = 3.47
pH = 14-pOH = 14-3.47 = 10.53
e) Select an appropriate acid/base indicator to find the end point for this titration. Predict the color change that will be observed at the end point.
Thymolphtalein, will change about pH = 10 from transparent to blue