Question

A 40.00 mL sample of 0.10 M weak acid with Ka of 1.8×10−5 is titrated with...

A 40.00 mL sample of 0.10 M weak acid with Ka of 1.8×10−5 is titrated with a 0.10 M strong base. What is the pH after 20.00 mL of base has been added?

Homework Answers

Answer #1

this is a weak acid + strong base

addition of base will form conjugate of the weak acid i.e. conjugate base

this will be a buffer

Apply buffer equation

pH = pKa + log(conjujgate/acid)

pKa = -log(ka) = -log(1.8*10^-5) = 4.74472

Total volume = V1+V2 = 40 + 20 = 60 ml

mol of base added = MV = 0.1*20 = 2 mmol of base added

mol of conjugate formed = 0 + 2 mmol

mol of acid initially = 40*0.1= 4 mmol acid

mol of acid left = 4-2= 2 mmol left

note that this is the HALF equivalence point

this is a pretty unique step since

[Conjugate] = [Acid]

log(1) = 0

then

pH = pKa + log(conjujgate/acid)

pH = pKa = 4.747

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate...
A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate the pH of the solution during the titration if the weak acid concentration is 0.10 M and its Ka = 1.8 x 10-5 and 10.0 mL of base has been added.
50.0 mL of 0.10 M HA, a weak acid (Ka= 1.5 × 10-6), is titrated with...
50.0 mL of 0.10 M HA, a weak acid (Ka= 1.5 × 10-6), is titrated with 0.10 M B, a strong base. What is the pH at Vb= ½ Ve= 25.0 mL?
30.0 mL sample of 0.10 M CH3COOH is titrated with 0.12 M NaOH. Determine the pH...
30.0 mL sample of 0.10 M CH3COOH is titrated with 0.12 M NaOH. Determine the pH of the solution; a) Before the addition of the base. The Ka of CH3COOH is 1.8 × 10-5. (5 points) b) After the addition of 25.0 mL of NaOH. The Ka of CH3COOH is 1.8 × 10-5. (5 points)
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a...
A 25.0 mL sample of a 0.115 M solution of acetic acid is titrated with a 0.144 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. (The Ka for acetic acid is 1.76 x 10^-5). 10.0 mL of base = 20.0 mL of base = 30.0 mL of base =
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. More...
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 40.0 mL ?
1. 50.0 ml of 0.10 M HNO2 (Ka = 4.0 x 10-4) are being titrated with...
1. 50.0 ml of 0.10 M HNO2 (Ka = 4.0 x 10-4) are being titrated with 0.10 M NaOH. The pH after 25.0 ml of NaOH have been added is...? 2. If 25 ml of 0.75M HCl are added to 100 ml of 0.25M NaOH, what is the final pH? thank you!
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. 1)A...
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. 1)A solution is made by titrating 9.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH? 2)More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 41.0 mL ?
A 32.4 mL sample of a 0.540 M aqueous hypochlorous acid solution is titrated with a...
A 32.4 mL sample of a 0.540 M aqueous hypochlorous acid solution is titrated with a 0.317 M aqueous barium hydroxide solution. What is the pH after18.7 mL of base have been added? Design a buffer that has a pH of 8.56 using one of the weak base/conjugate acid systems shown below. Weak Base Kb Conjugate Acid Ka pKa CH3NH2 4.2×10-4 CH3NH3+ 2.4×10-11 10.62 C6H15O3N 5.9×10-7 C6H15O3NH+ 1.7×10-8 7.77 C5H5N 1.5×10-9 C5H5NH+ 6.7×10-6 5.17 How many grams of the chloride...
A 50.00 ml aliquot of 0.1000 M benzoic acid (Ka= 6.28x10^-5) was titrated with 0.2500 M...
A 50.00 ml aliquot of 0.1000 M benzoic acid (Ka= 6.28x10^-5) was titrated with 0.2500 M KOH. Calculate the pH of the solution after the addition of O.00, 5.00, 10.00,15.00,19.95, 20.00, 20.05, 20.50,and 25.00ml of base.
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M...
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M HCl. If pK_bb​ = 4.76 for ammonia, what is the solution pH? 2) The single equivalence point in a titration occurs at pH = 5.65. What kind of titration is this? A Strong acid titrated with a strong base B Weak acid titrated with a strong base C Weak base titrated with a strong acid
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT