Question

20 mL of 0.5 M NaCN solution is titrated with 0.4 M HCl solution. Calculate the...

  1. 20 mL of 0.5 M NaCN solution is titrated with 0.4 M HCl solution. Calculate the pH after the addition of:
  1. 10 mL HCl
  2.   25 mL HCl

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 46.3 mL sample of a 0.380 M solution of NaCN is titrated by 0.350 M...
A 46.3 mL sample of a 0.380 M solution of NaCN is titrated by 0.350 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of the solution: (a) prior to the start of the titration pH = (b) after the addition of 25.1 mL of 0.350 M HCl pH = (c) at the equivalence point pH = (d) after the addition of 71.9 mL of 0.350 M HCl. pH =
A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M...
A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of the solution: (a) prior to the start of the titration pH = (b) after the addition of 48.4 mL of 0.260 M HCl pH = (c) at the equivalence point pH = (d) after the addition of 76.2 mL of 0.260 M HCl. pH =
20 mL of 0.25 M of NH3 is titrated with 0.40 M HCl. Calculate the pH...
20 mL of 0.25 M of NH3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. Kb NH3 = 1.8 × 10−5
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
A 40.0 mL volume of 0.100 M NaOH is titrated with 0.0500 M HCl. Calculate the...
A 40.0 mL volume of 0.100 M NaOH is titrated with 0.0500 M HCl. Calculate the pH after addition of the following volumes of acid. 1.) 64 mL 2.) 89.4 mL 3.) 92 mL
10 mL of 0.10 M HCl is being titrated with 0.10 M NaOH. What would the...
10 mL of 0.10 M HCl is being titrated with 0.10 M NaOH. What would the pH of the solution be after addition of a) 1.0 mL of NaOH and b) 9.0 mL of NaOH
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...
a) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217...
a) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is the pH after 75 ml of HCl have been added? b) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is the pH after 25 ml of HCl have been added? c) 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCl. What is...
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH...
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 25 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.
A volume of 25.0 mL OF 0.140 M HCl is titrated against a 0.140 M CH3NH2...
A volume of 25.0 mL OF 0.140 M HCl is titrated against a 0.140 M CH3NH2 solution added to it from a buret. A.) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added. B.) Calculate the pH value of the solution after 25.0 mL of CH3NH2 solution have been added. C.) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.