6 A 300mL sample of 0165 M propanoic acid K 13 x 10 5 is tit

6. A 30.0-mL sample of 0.165 M propanoic acid (K, 1.3 x 10 5) is titrated with 0.300 M KOH. Calculate the pH at: a) 0 mL added base b) 10 mL added base c) equivalence point d) half-equivalence point e) 20 mL added base

Solution

Propanoic acid= 30.0 ml of 0.165M

Number of moles of Propanoic acid= 0.165M x 0.030L= 0.00495moles

a) 0 ml of base

for weak acids

[H+] = square root of KaXC

[H+] = square root of (1.3x10^-5 x 0.165)

[H+] = 1.465x10^-3M

-log[H+] = -log[1.465x10^-3]

PH= 2.83

b) 10 ml of base

KOH= 10 ml of 0.300M

number of moles of KOH= 0.300M x0.010L= 0.003 moles

CH3-CH2-COOH + KOH ---------------- CH3-CH2-COOK   + H2O

0.00495                  0.003                        0

- 0.003                   - 0.003                 + 0.003

after additon of KOH

remaining number of moles of Propanoic acid = 0.00495 -0.003= 0.00195 moles

number of moles of salt = 0.003 moles

Ka= 1.3x10^-5

-log(Ka) = -log(1.3x10^-5)

PKa = 4.89

PH= PKa + log[salt]/[acid]

PH= 4.89 + log[0.003/0.00195]

PH= 5.08

c)equivalne point

at equivalent point

number of moles of acid = number of moles of base

number of moles = 0.00495 moles

volume of KOH= 0.165 x30.0/0.30 = 16.5 ml

total volume of solution= 30.00+ 16.5 = 46.5 ml = 0.0465L

C= 0.00495/0.0465 = 0.106M

At equivalent

PH= 7 + 1/2{PKa + logC}

PH= 7 + 1/2{4.89 + log(0.106)}

PH= 8.957

PH= 8.96

d) half equivalent point

PH= Pka

PH= 4.89

e) at 20 ml of base

number of moles ofKOH = 0.300M x0.020L= 0.006 moles

number o fmoles of base is greaterthan the acid

remaining number o fmoles of base = 0.006 - 0.00495= 0.00105 moles

total volume= 30.00+ 20.0= 50.0 ml= 0.050L

[OH-] = 0.00105/0.050=0.021M

[OH-] = 0.021M

-log{OH-} = -log[0.021]

POH= 1.678

PH + POH= 14

PH= 14 - POH

PH= 14- 1.678

PH= 12.32