Question

what are the last three values?

Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.2400 mol/L butanoic acid, CHjCH2CH2COOH (Ka 1.54 x 10-5), with 0.2400 mo/L NaOH solution after each addition of titrant: (a) 0 mL (b) 10.00 ml. pH= 4.81 (c) 15.00 mL pH = 5.29 (d) 19.00 ml. pH = 6.08 (e) 19.95 mL pH = 7.41 (f) 20.00 mL pH= (g) 20.05 mLpH- (h) 25.00 mLpH- pH = 2.71

Answer 1

CH3-CH2-CH2-COOH = 20.00mL of 0.240 mole

number of moles of CH3-CH2-CH2-COOH = 0.240M x 0.02000L= 0.0048 mole

Ka= 1.54 x10^-5

NaOH= 0.240 M

f) volume of NaOH = 20.00 mL

number of moles fo NaOH = 0.240M x 0.0200L = 0.0048 mole

number of moles of CH3-CH2-CH2-COOH is equal to NaOH

so it is the equivalent point

at equivalent point

PH= 7+1/2[PKa + logC]

Total volume = 20.00+20.00= 40.00ml = 0.0400L

C= number of moles/volume = 0.0048/0.040= 0.12M

Ka= 1.54 x 10^-5

-log(Ka) = -log(1.54x10^-5)

PKa= 4.81

PH = 7 + 1/2[ 4.81 + log(0.12)]

PH= 8.94

g) Volume of NaOH = 20.05 mL

number of moles of NaOH = 0.240M x 0.02005L = 0.004812 mole

number of moles of CH3-CH2-CH2-COOH = 0.0048 mole

number of moles of NaOH is greater than the number of moles of CH3-CH2-CH2-COOH

so the nature of the solution is basic

remaining number of moles of NaOH = 0.004812 - 0.0048 = 0.000012 mole

Total volume = 20.00+20.02= 40.02ml = 0.04002L

[OH-] = number of moles/volume = 0.000012/0.04002= 2.998 x 10^-4M

[OH-] = 2.998 x 10^-4

-log[OH-] = -log(2.998x10^-4)

POH= 3.52

PH + POH = 14

PH = 14 -POH

PH = 14 - 3.52

PH = 10.48

h) volume of NaOH = 25.00mL

Number of moles of NaOH = 0.240M x 0.0250L = 0.006 mole

number of moles of CH3-CH2-CH2-COOH = 0.0048 mole

so the nature of the solution is basic

remaining number of moles of NaOH = 0.006 - 0.0048 = 0.0012 moles

Total volume = 20.00 + 25.00= 45.00mL = 0.045L

[OH-] = 0.0012/0.045= 0.0267

[OH-] =0.0267

-log[OH-] = -log(0.0267)

POH= 1.57

PH= 14 - 1.57

PH= 12.43.