Question

7a. What is the pH at 0.00 mL of titrant in the titration of 60.00 mL of 0.200 M HA (a generic acid with K_a = (2.59x10^-5))with 0.400 M NaOH?

7b. What is the pH at 25.00 mL of titrant in the titration of 60.00 mL of 0.200 M HA (a generic acid with K_a = (2.11x10^-5))with 0.400 M NaOH?

Please help, will rate.

Answer 1

7a)

Generc acid =HA = 60.00 mL of 0.200M

Ka = 2.59 x10^-5

for weak acids

concentratio of H+ = square root of ( Ka x C)

[H+] = square root of ( 2.59 x10^-5 x 0.200)

[H+] = 2.28 x10^-3 M

-log [H+] = - log( 2.28 x10^-3)

PH = 2.64

7b)

Generic acid = HA = 60.00 mL of 0.200M

number of moles = Molarity x volume in L

number of moles of HA = 0.200 M x 0.0600L = 0.0120 moles

NaOH = 25.00 mL of 0.400M

number of moles of NaOH = 0.400 M x 0.0250L = 0.010 moles

Ka = 2.11 x 10^-5

-log(Ka) = - log( 2.11 x10^-5)

PKa = 4.67

HA +_ NaOH ----------------- NaA + H2O

Initial 0.0120 0.010 0

change - 0.010 - 0.010 + 0.010

equilibrium 0.002 0 + 0.010

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

PH = 4.67 + log( 0.010/ 0.002)

PH = 5.37