Question

What is the pH when 5.0 g of sodium acetate, NaC2H3O2, is dissolved in 150.0 mL...

What is the pH when 5.0 g of sodium acetate, NaC2H3O2, is dissolved in 150.0 mL of water? (The Ka of acetic acid, HC2H3O2, is 1.8×10−5.)

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
1. Determine the pH when 5.89 g of sodium acetate (NaOOCCH3) is dissolved in 190 mL...
1. Determine the pH when 5.89 g of sodium acetate (NaOOCCH3) is dissolved in 190 mL of water. Ka of acetic acid is 1.8 10-5                                     NaOOCCH3Na+   +  -OOCCH3                                     -OOCCH3  +  H2OHOOCCH3  +   OH- 2. A student wants to kill Dr. Dahm, so he walks buys 10 grams of potassium cyanate (KCN) and drops all 10 grams into Dr. Dahms pepsi can.  Determine the pH if the can holds 335 mL of liquid and the Ka of HCN is 6.2 10-10.                                     KCN   K+ +  CN-                                     CN-   +  H2O    HCN +  OH-
One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations...
One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient....
Calculate the pH of a solution of 0.100M HC2H3O2 (acetic acid) and 0.125M NaC2H3O2 (sodium acetate)...
Calculate the pH of a solution of 0.100M HC2H3O2 (acetic acid) and 0.125M NaC2H3O2 (sodium acetate) using an ICE (ICF) table. Calculate the pH of a solution of 0.010M C5H5NH+Cl- (pyridinium hydrochloride) and 0.006M C5H5N (pyridine) using an ICE (ICF) table.
You will determine the amount of sodium acetate needed to prepare 50.0 mL of an acetic...
You will determine the amount of sodium acetate needed to prepare 50.0 mL of an acetic acid/sodium acetate buffer having a pH of 5.0 using approximately 0.100M acetic acid. For acetic acid Ka = 5.6 x 10^-5. How do I determine the sodium acetate needed? Thank you.
Find the pH of a 0.342 M NaC2H3O2 solution. (The Ka of acetic acid, HC2H3O2, is...
Find the pH of a 0.342 M NaC2H3O2 solution. (The Ka of acetic acid, HC2H3O2, is 1.8×10−5.)
How many grams of sodium acetate must be added to a 150.0 mL solution of 0.40...
How many grams of sodium acetate must be added to a 150.0 mL solution of 0.40 M acetic acid to prepare a buffer with a pH = 4.80?
What is the pH of a 1.288 M solution of sodium acetate? Ka of acetic acid...
What is the pH of a 1.288 M solution of sodium acetate? Ka of acetic acid is 1.8*10^-5. (2 decimal places)
You will prepare 50.0 mL of a pH = 5.0 acetic acid/sodium acetate buffer solution. You...
You will prepare 50.0 mL of a pH = 5.0 acetic acid/sodium acetate buffer solution. You will use 10.0 mL of 0.100 M acetic acid. You will need to determine the amount of sodium acetate (the conjugate base) to use to prepare the buffer. How do I determine the amount of sodium acetate needed for my buffer?
Acetic acid (molecular weight = 60.05 g/mol) and sodium acetate (molecular weight = 82.03 g/mol) can...
Acetic acid (molecular weight = 60.05 g/mol) and sodium acetate (molecular weight = 82.03 g/mol) can form a buffer with an acidic pH. In a 500.0 mL volumetric flask, the following components were added together and mixed well, and then diluted to the 500.0 mL mark: 100.0 mL of 0.300 M acetic acid, 1.00 g of sodium acetate, and 0.16 g of solid NaOH (molecular weight = 40.00 g/mol). What is the final pH? The Ka value for acetic acid...
Calculate the pH after 0.018 mole of HCl is added to 1.00 L of each of...
Calculate the pH after 0.018 mole of HCl is added to 1.00 L of each of the three solutions. (Assume that all solutions are at 25°C.) a) 0.136 M acetic acid (HC2H3O2, Ka = 1.8 ✕ 10−5) b) 0.136 M sodium acetate (NaC2H3O2) c)  0.136 M HC2H3O2 and 0.136 M NaC2H3O2
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT