Question

calculate the volume of 1.50 × 10-2 M NaOH that must be added to 500.0 mL...

calculate the volume of 1.50 × 10-2 M NaOH that must be added to 500.0 mL of 0.200 M HCl to give a solution that has pH = 2.05

.Volume = __L

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

      PH = 2.05

   -log[H^+] = 2.05

       [H^+]   = 10^-2.05   = 0.0089M    it is resultant molarity

    HCl                     NaOH

MA = 0.2M    MB = 1.5*10^-2M = 0.015M

VA = 500ml VB =

M    = MAVA- MBVB/VA+VB

0.0089 = 0.2*500-0.015*VB/500+VB

0.0089*(500+VB) = 0.2*500-0.015*VB

VB = 3997.9ml

The volume of NaOH = 3998ml = 3.998L >>>>answer

                         

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A volume of 500.0 mL of 0.100 M NaOH is added to 535 mL of 0.200...
A volume of 500.0 mL of 0.100 M NaOH is added to 535 mL of 0.200 M weak acid (?a=4.69×10^−5). What is the pH of the resulting buffer? HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
Calculate the volume (in mL) of 0.170 M NaOH that must be added to 345 mL...
Calculate the volume (in mL) of 0.170 M NaOH that must be added to 345 mL of 0.0499 M 3-(N-Morpholino)propanesulfonic acid (MOPS) to give the solution a pH of 7.55. The pKa of MOPS = 7.18.
500.0 mL of 0.140 M NaOH is added to 615 mL of 0.200 M weak acid...
500.0 mL of 0.140 M NaOH is added to 615 mL of 0.200 M weak acid (Ka = 4.06 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.110 M NaOH is added to 585 mL of 0.200 M weak acid...
500.0 mL of 0.110 M NaOH is added to 585 mL of 0.200 M weak acid (Ka = 7.30 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.130 M NaOH is added to 575 mL of 0.200 M weak acid...
500.0 mL of 0.130 M NaOH is added to 575 mL of 0.200 M weak acid (Ka = 7.35 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.170 M NaOH is added to 575 mL of 0.250 M weak acid...
500.0 mL of 0.170 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 1.50 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.100 M NaOH is added to 575 mL of 0.250 M weak acid...
500.0 mL of 0.100 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 3.61 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.120 M NaOH is added to 605 mL of 0.250 M weak acid...
500.0 mL of 0.120 M NaOH is added to 605 mL of 0.250 M weak acid (Ka = 4.44 × 10-5). What is the pH of the resulting buffer?
500.0 mL of 0.160 M NaOH is added to 575 mL of 0.250 M weak acid...
500.0 mL of 0.160 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 2.43 × 10-5). What is the pH of the resulting buffer?
1.What volume of a 1.50 M NaOH solution should be added to 50.0 mL of 1.20...
1.What volume of a 1.50 M NaOH solution should be added to 50.0 mL of 1.20 M acetic acid (CH3CO2H; Ka = 1.76×10-5) to obtain a buffer with pH = 5.450? a) 33.3 mL b) 6.70 mL c) 16.4 mL d) 67.0 mL e) 42.1 mL 2. Find the concentration of hydronium ion, H3O+, in a 0.200 M solution of sodium hypochlorite, NaOCl. Ka(HOCl) = 3.0×10-8. a) 7.8×10-5 M b) 1.3×10-10 M c) 1.0×10-7 M d) 3.9×10-11 M e) 2.6×10-4...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT