Question

1. a. What is the pH of a solution formed by combining 50.00 mL of a 0.0855 M acetic acid solution, 25.00 mL of 0.1061 M NaOH solution and 25.00 mL of deionized water? Show your work.

b. What is the pH when the solution in 1a is diluted 1 mL in 10 mL total volume? Show your work.

c. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M HCl solution added to it? Show your work.

c. What is the pH when 25.00 mL of the solution in #1a has 5.00 mL of a 0.1056 M NaOH solution added to it? Show your work.

Answer 1

1 a. no of mole of aceticacid = M*V

                             = 0.0855*50 = 4.275 mmole

    no of mole of NaOH = 0.1061*25 = 2.6525 mmole

   pka of aceticacid = 4.74

pH = pka + log(base/acid)

     = 4.74 + log(2.6525/(4.275-2.6525))

     = 4.953

b. no of mole of aceticacid in diluted solution = 4.275*(1/75)*(1/10)

                       = 0.0057 mmole

    no of mole of NaOH = 2.6525*(1/75)*(1/10) = 0.00353 mmole

   pka of aceticacid = 4.74

pH = pka + log(base/acid)

     = 4.74 + log(0.00353/(0.0057-0.00353))

     = 4.951

c. after addition of HCl

no of moleof HCl added = 5*0.1056 = 0.528 mmole

no of mole of aceticacid in 25 ml solution = 4.275*(25/75) = 1.425 mmole

no of mole of NaOH in 25 ml solution = 2.6525*(25/75) = 0.88 mmole

pka of aceticacid = 4.74

pH = pka + log(base-acid/acid+acid)

     = 4.74 + log((0.88-0.528)/(1.425+0.528))

     = 4.0

d.   after addition of NaOH

no of moleof NaOH added = 5*0.1056 = 0.528 mmole

no of mole of aceticacid in 25 ml solution = 4.275*(25/75) = 1.425 mmole

no of mole of NaOH in 25 ml solution = 2.6525*(25/75) = 0.88 mmole

pka of aceticacid = 4.74

pH = pka + log(base+NaOH/acid-NaOH)

     = 4.74 + log((0.88+0.528)/(1.425-0.528))

     = 4.936