Question

When 100.0 mL of a weak acid was titrated with 0.09381 M NaOH, 27.63 mL were...

When 100.0 mL of a weak acid was titrated with 0.09381 M NaOH, 27.63 mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What was the pH when only 19.47 mL of NaOH had been added?

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

millimoles of NaOH added = 0.09381 x 27.63 = 2.592

2.592 millimoles acid must be present.

2.592 = M x 100.0

M = 0.02592 M

[acid] = 0.02592 M

after equivalence point reached

[salt] = 2.592 / 100 + 27.63 = 0.0203 M

pH = 1/2 [pKw + pKa + logC]

10.99 = 1/2 [14 + pKa + log 0.0203]

21.98 = pKa + 12.31

pKa = 9.67

millimoles of NaOH added = 19.47 x 0.09381 = 1.826

2.592 - 1.826 = 0.766 millimoles acid left

1.826 millimoles salt formed

[acid] = 0.766 / 100 + 19.47 = 0.0064 M

[salt] = 1.826 / 119.47 = 0.0153 M

pH = pKa + log [salt] / [acid]

pH = 9.67 + log [0.0153] / [0.0064]

pH = 10.05

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
When 100.0 mL of weak acid HA were titrated with 0.09381 M RbOH, 27.36 mL were...
When 100.0 mL of weak acid HA were titrated with 0.09381 M RbOH, 27.36 mL were required to reach the equivalence point. (a) Find the molarity of HA. (b) What is the formal concentration of A- at the equivalence point? (c) The pH at the equivalence point was 10.99. Find pKa for HA. (d) What was the pH when only 19.47 mL of RbOH has been added?
When 10.00 ml of NH3 solution is titrated with 0.015 M Hbr, 27.5ml of the acid...
When 10.00 ml of NH3 solution is titrated with 0.015 M Hbr, 27.5ml of the acid is needed to reach the equivalence point. what is the concentration of the ammonia? what is the pH when 10.00mL 0.015 M HBr has been added and what is the equivalence point pH?
A 100-mL solution of weak acid was titrated with 0.09145 M NaOH solution. The titration showed...
A 100-mL solution of weak acid was titrated with 0.09145 M NaOH solution. The titration showed that Ve/4 = 8.55 mL, and the measured pH at that volume was 5.00. Determine the pH at the equivalence point, and choose an appropriate endpoint indicator. Hint: recall that the H-H equation does not apply at the equivalence point.
A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH...
A 25.0 mL sample of an acetic acid solution in titrated with a 0.09984 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. Calculate the concentration of acetic acid in the 25.0 mL sample. Calculate the pH at the equivalence point (Ka acetic acid = 1.75x10^-5)
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH....
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH. To reach the endpoint of the titration, 30.00 mL of NaOH solution is required. Ka = 1.8 x 10-4 What is the pH of the solution after the addition of 10.00 mL of NaOH solution? What is the pH at the midpoint of the titration? What is the pH at the equivalence point?
When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following...
When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration? A. Initially the pH will be less than 1.00. B. The pH at the equivalence point will be 7.00. C. It will require 12.50 mL of NaOH to reach the equivalence point. When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration? A. Initially the pH will be...
Question1: A 100.0 mL solution of 0.5 M histidine in its tribasic form, A3-, (pKa1 =...
Question1: A 100.0 mL solution of 0.5 M histidine in its tribasic form, A3-, (pKa1 = 1.70, pKa2 = 6.02, pKa3 = 9.08) is titrated with 1.0 M HCl titrant. 1)calculate the volume of titrant required to reach the second equivalence point, 2)calculate the pH after the addition of 50.0 mL of titrant, 3)calculate the pH after the addition of 62.0 mL of titrant, 4)calculate the pH after the addition of 75.0 mL of titrant. Question4: A 100 mL volume...
25.000 mL of 0.1114 M propanoic acid solution is titrated with 0.1308 M NaOH. Calculate the...
25.000 mL of 0.1114 M propanoic acid solution is titrated with 0.1308 M NaOH. Calculate the pH of titrant mixture at the following volumes of NaOH added a) 12.72 mL b) 21.29 mL (Equivalence point)
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 ?
2. The Ka of nitrous acid is 4.5 x 10-4. In a titration 50.0 mL of...
2. The Ka of nitrous acid is 4.5 x 10-4. In a titration 50.0 mL of a 1.00 M solution is titrated with 0.750 M NaOH. Calculate the pH of the solution: a. Before the titration begins. b. When sufficient NaOH has been added to neutralize half of the nitrous acid originally present. c. At the equivalence point. d. When 0.05 mL of NaOH less than that required to reach the equivalence point has been added. e. When 0.05 mL...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT