Question

Calculate the pH of a solution prepared by mixing 15.0 mL of 0.10 M NaOH and 30.0 mL of a 0.10 M benzoic acid solution. (Benzoic acid is monprotic; its dissociation constant is 6.5 x 10^-5.)

Answer 1

Balance Equation
C65COOH(aq) +OH(aq) -> C6H5COO+ Na(aq) +H20(l)

1.) Get units of Molar Concentration
Moles NaOH= ( 0.015L)(0.10) = 0.0015M NaOH
Moles C6H5COOH= ( 0.030L)( 0.10)= 0.030M C6H5COOH Benzoic Acid

2.) Find total concentration of C6H5COOH, this will determine if this solution is a buffer
C6H5COOH= 0.0015mol/ 0.0450L = 0.0333M
C6H5COO = 0.0015mol/ 0.00450L= 0.0333M
since they are equal concentrations, this means this is a Buffer

3.) PH=Pka+log Benzoic acid/ Benzoic Ion

so you can just do this
pH= -log( 6.5*10^-5)
pH= 4.19

Answer 2

Let me know if you understand

Answer 3

@n648c788 i understand but doesn't (0.030L)(0.10)= 0.0030M NOT 0.030M ?

Answer 4

yes I forgot an extra o sorry, but that is how you do it

Answer 5

ok, so the answer would not be 4.19?

Answer 6

It will be 4.19, because the extra work is to prove it is a buffer, so just take the -log of the 6.5*10^-5

Answer 7

If I calculated part 2 wrong then we would have a different answer, but I did part 2 correctly it showed that this reaction is a buffer, so we were able to take the -log of the ka

Answer 8

Just notfity I, if you still need clarification the answer is 4.19

Answer 9

pH= pka +log conj acid/base

pka=-log

pH=-log(6.5*10^-5)+log(0.0333M/0.0333M)

pH=4.19+0

pH=4.19