Question

Refer to the equilibrium constants. Calculate the pH of a solution prepared by adding 60.0 mL of 0.100 M NaOH to 100. mL of 0.100 M CH3COOH solution (CH3COOH Ka = 1.8*10^-5) . Answer is B, please show work.

A)4.56

B)4.92

C)5.00

D)5.08

E)5.16

Answer 1

NaOH = 60.0mL of 0.100M

number of moles of NaOH = 0.100M x 0.060L= 0.006 moles

CH3COOH = 100 mL of 0.100M

number of moles of CH3COOH = 0.100Mx 0.100L = 0.01 moles

Ka = 1.8x10^-5

-log(Ka) = -log( 1.8x10^-5)

PKa= 4.74

                   CH3COOH   + NaOH -------------- CH3COONa    + H2O

                     0.01                 0.006                            0

                 - 0.006               - 0.006                     + 0.006

                 0.004                       0                          + 0.006

PH = PKa +log[salt/acid]

PH = 4.74 + log(0.006/0.004)

PH = 4.916

PH = 4.92

The answer is B.