Question

Kathy weighed out 6.20 grams of sodium benzoate, NaC7H5O2 along with 2.25 grams of benzoic acid,...

Kathy weighed out 6.20 grams of sodium benzoate, NaC7H5O2 along with 2.25 grams of benzoic acid, HC7H5O2, and dissolved it together in 750.0 mL of water.

What is the pH of this solution if Kathy adds 1.50 mL of 1.0 M HCl to 20.00 mL of this solution? (no clue on how to do this, please show me)

Homework Answers

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 250mL buffer is made of .33 M benzoic acid and .21 M sodium benzoate. What...
A 250mL buffer is made of .33 M benzoic acid and .21 M sodium benzoate. What is the initial ph of the solution? If 1.0 mL of 12M NaOH is added to the solution, what is the new ph?
What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that...
What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.56 and has a freezing point of -2.0 ∘C? Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)
what mass of sodium benzoate must be dissolved in 500 ml of .65 M benzoic acid...
what mass of sodium benzoate must be dissolved in 500 ml of .65 M benzoic acid to produce a buffer with the pH of 4.7
What mass of sodium benzoate should be added to 150.0 mL of a 0.17 M benzoic...
What mass of sodium benzoate should be added to 150.0 mL of a 0.17 M benzoic acid solution in order to obtain a buffer with a pH of 4.30? (answer is in grams).
common methods to prepare a buffer. Calculate the mass of benzoic acid and sodium benzoate (in...
common methods to prepare a buffer. Calculate the mass of benzoic acid and sodium benzoate (in grams) needed to prepare 250 mL of a 0.1 M buffer at pH = 4.1. (The buffer concentration is defined as the sum of the conjugate acid concentration plus the conjugate base concentration.) The pKa of benzoic acid is 4.2
A 250mL buffer is made of .33 M benzoic acid and .21 M sodium benzoate. What...
A 250mL buffer is made of .33 M benzoic acid and .21 M sodium benzoate. What is the initial pH of the solution? If 1.0 mL of 12M NaOH is added to the solution, what is the new pH? Then if the concentrations in part A of this experiment had been 0.50 M for the acid and 1.5 M for the base, what would the pka have been, assuming the pH remains the same?
What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that...
What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.78 and has a freezing point of -2.0 ∘C? (Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.) Express your answers using two significant figures separated by a comma. answer is not 0.11 and 0.44 answer is not 0.11 and 0.45 answer is not 0.11 and 0.46 Textbook says Tf is 1.86
Calculate the pH of a solution prepared by dissolving 0.170 mol of benzoic acid and 0.310...
Calculate the pH of a solution prepared by dissolving 0.170 mol of benzoic acid and 0.310 mol of sodium benzoate in water sufficient to yield 1.50 L of solution. The Ka of benzoic acid is 6.30 × 10-5. Please show steps to help me understand this.
Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is...
Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water: C6H5COOH⇌C6H5COO−+H+ The pKa of this reaction is 4.2. In a 0.61 M solution of benzoic acid, what percentage of the molecules are ionized?
What mass of sodium benzoate should be added to 140.0 mL of a 0.13 M benzoic...
What mass of sodium benzoate should be added to 140.0 mL of a 0.13 M benzoic acid solution in order to obtain a buffer with a pH of 4.25? Please help!! I dont understand the process
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT