Question

1. When 50.00 mL of 0.100 M NaOH reacts with 25.00 mL of 0.300 M HX,...

1. When 50.00 mL of 0.100 M NaOH reacts with 25.00 mL of 0.300 M HX, a weak monoprotic acid, the pH of the resulting solution is 3.74. Calculate the Ka of the acid.

2. Will the equivalence point of ammonia plus HCl be higher, lower, or equal to 7? Explain.

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

1) [NaOH ] = molarity x volume in Litres = 0.1 M x 0.05 L = 0.005 mol

[HX] = 0.3 M x 0.025 L = 0.0075 mol

HX + NaOH ---------------> NaX + H2O

0.0075 mol 0.005 mol 0

-----------------------------------------------------------------------------

0.0075 - 0.005 0 0.005 mol

= 0.0025 mol

Hence,

[HX] = 0.0025 mol

[NaX] = 0.005 mol

pH = -logKa + log [NaX] /[HX]

3.74 = -logKa + log [0.005]/[0.0025]

logKa = log [0.005]/[0.0025] - 3.74

Hence,

Ka = 0.000364

Therefore,

Ka of the acid =  0.000364

2) Equivalence point of ammonia plus HCl

  It will be lower than 7.

Because, HCl is a strong acid and ammonia is weak base.

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 25.0 mL of 0.100 M NaOH was added to 50.0 mL of a 0.100 M...
When 25.0 mL of 0.100 M NaOH was added to 50.0 mL of a 0.100 M solution of a weak acid, HX, the pH of the mixture reached a value of 3.56. What is the value of Ka for the weak acid?
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...
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M...
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M HCl. If pK_bb​ = 4.76 for ammonia, what is the solution pH? 2) The single equivalence point in a titration occurs at pH = 5.65. What kind of titration is this? A Strong acid titrated with a strong base B Weak acid titrated with a strong base C Weak base titrated with a strong acid
A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with 0.100 NaOH....
A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with 0.100 NaOH. a) the initial pH (before addition of any NaOH) is? b)What is the pH at the first half-equivalence point (halfway to the first equivalence point)? c)What is the pH at the first equivalence point? d)What is the pH at the second half-equivalence (halfway between the first and second equivalences)? e)What is the pH at the second equivalence point? f)What is the pH after adding...
What is the pH at the equivalence point when 55.0 mL of a 0.300 M solution...
What is the pH at the equivalence point when 55.0 mL of a 0.300 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate the concentration of the acid in the original solution, the pH of the original HCl solution and the original NaOH solution
Consider the titration of 25.00 mL of 0.300 M H3 PO4 with 0.300 M NaOH. 1.What...
Consider the titration of 25.00 mL of 0.300 M H3 PO4 with 0.300 M NaOH. 1.What is the pH of the titration medium after the addition of the following volumes of NaOH titrant: F. 50.00 mL: G. 60.00 mL: H. 62.50 mL: I. 75.00 mL: J. 85.00 mL Can writer should how to find Ka1,Ka2, and Ka3?
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?
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?
1. When 4.00 mL of 0.100 M HCl is added to each solution, what is the...
1. When 4.00 mL of 0.100 M HCl is added to each solution, what is the pH? We have not seen an example of each of these—reason it out. Argue with your study group. Think through the flowchart. Ka of HF is 2.4 x 10-4 a. 250.00 mL that contains both 0.100 M HF and 0.125 M KF b. 50.00 mL that contains both 0.080 M NH3 and 0.090 M NH4Cl c. 50.00 mL that contains only 0.100 M NaOH...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT