14) you are titrating 10.0mL of a solution of 0.25M acetic acid (Ka=1.8 x 10 ^-5)....

14) you are titrating 10.0mL of a solution of 0.25M acetic acid (Ka=1.8 x 10 ^-5). you are using a solution of 0.10M KOH to complete the titration curve. create the graph that would result by adding 1.0mL of the base to the acid and determining the pH and repeating until you have passed the equivalence point. Start the graph with 0 mL of the base solution added.

Expert Solution

queen_honey_blossom answered 1 year ago

a. The ka of acetic acid is 1.8*10^-5. What is the pka of the acetic acid?...

a. The ka of acetic acid is 1.8*10^-5. What is the pka of the acetic acid? b.What should be the ph of a solution of 1.0 mL of 0.1 M acetic acid and 1.0 mL of 0.1 M sodium acetate and 48.0 mL H2O? c.What should be the ph of a solution of 5.0 ml of 0.1 M acetic acid and 5.0ml of 0.1M sodium acetate and 40.0ml h20? is the answer 4.74 for all parts ?? thanks

The Ka of acetic acid, CH3COOH, is 1.8 × 10-5. A buffer solution was made using...

The Ka of acetic acid, CH3COOH, is 1.8 × 10-5. A buffer solution was made using an unspecified amount of acetic acid and 0.30 moles of NaOOCCH3 in enough water to make 1.50 L of solution. The pH of the solution is 4.55. How many moles of CH3COOH were used? Could you please showing working on how to get the answer of 0.47.

Given Ka for acetic acid = 1.8 x 10-5, calculate the pH in the titration of...

Given Ka for acetic acid = 1.8 x 10-5, calculate the pH in the titration of 50.0 mL of 0.120 M acetic acid by 0.240 M sodium hydroxide after the addition of the following: 0.00 mL of the base. 10.0 mL of base. 25.0 mL of the base. 35.0 mL of the base. Please show all work

Calculate the pH of a 0.51 M CH3COOK solution. (Ka for acetic acid = 1.8×10−5.)

Calculate the pH of a 0.800M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8×10^-5.

25.00 mL of acetic acid (Ka = 1.8 x 10-5) is titrated with a 0.09991M NaOH.

1) 25.00 mL of acetic acid (Ka = 1.8 x 10-5) is titrated with a 0.09991M NaOH. It takes 49.50 mL of base to fully neutralize the acid and reach equivalence. What is the concentration of the acid sample? What is the initial pH? What is the pH after 20.00 mL of base has been added? What is the pH at equivalence ? What is the pH after 60.00 mL of base has been added? What indicator should be...

Weak Acid Ka Weak Base Kb CH3COOH (Acetic Acid) 1.8 X 10-5 NH3 (Ammonia) 1.76 X...

Weak Acid Ka Weak Base Kb CH3COOH (Acetic Acid) 1.8 X 10-5 NH3 (Ammonia) 1.76 X 10-5 C6H5COOH (Benzoic Acid) 6.5 X 10-5 C6H5NH2 (Aniline) 3.9 X 10-10 CH3CH2CH2COOH (Butanoic Acid) 1.5 X 10-5 (CH3CH2)2NH (Diethyl amine) 6.9 X 10-4 HCOOH (Formic Acid) 1.8 X 10-4 C5H5N (Pyridine) 1.7 X 10-9 HBrO (Hypobromous Acid) 2.8 X 10-9 CH3CH2NH2 (Ethyl amine) 5.6 X 10-4 HNO2 (Nitrous Acid) 4.6 X 10-4 (CH3)3N (Trimethyl amine) 6.4 X 10-5 HClO (Hypochlorous Acid) 2.9 X...

a 50 ml sample of 0.095 m acetic acid (ka=1.8 x 10^-5) is being titrated with...

a 50 ml sample of 0.095 m acetic acid (ka=1.8 x 10^-5) is being titrated with 0.106 M naoh. what is the pH at the equivalence point of the titration?

Using 1.8*10^-5 for the Ka for acetic acid, along with the weight of sodium acetate and...

Using 1.8*10^-5 for the Ka for acetic acid, along with the weight of sodium acetate and volume of acetic acid used to prepare a buffer, calculate the expected pH of the buffer. So the weight is 4.0089 g and the volume is 100mL

Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x...

Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x 10-5) (a) What is the pH of 5 mL of a 5.0% solution? (b) What is the pH of the solution if you now add 45 ml of water to solution (a)? How will the equivalence point volumes differ if you titrate the two solutions ?

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