1. a. What is the pH of a solution formed by combining 50.00 mL of a 0.0855 M ac
ID: 1053240 • Letter: 1
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.
Explanation / Answer
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
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