Question

In: Chemistry

A 40.00 mL aqueous solution of 0.100 M potassium cyanide is titrated with 0.150 M HBr....

A 40.00 mL aqueous solution of 0.100 M potassium cyanide is titrated with 0.150 M HBr. What is the pH of the solution at the equivalence point?
Correct Answer: 2.99
-I just want to know how its done. Thank you.

Solutions

Expert Solution

At equivalence point :

millimoles of CN- = millmoles of HBr

40 x 0.1 = 0.150 x V

V = 26.67 mL

CN- (aq)   + HBr    ---------------->   HCN   + Br-

4                4                                     0

0                 0                                     4

here HCN is formed .

concentration of [HCN] = 4 / (40 + 26.67) = 0.06 M

pKa = 9.21

pH = 1/2 (pKa - log C)

    = 1/2 (9.21 - log 0.06)

pH = 5.22

NOte : check your answer once. it is wrong. correct answer 5.22


Related Solutions

A 40.00 mL aliquot of 0.100 M NH3 is titrated with 20.00 mL of 0.0700 M...
A 40.00 mL aliquot of 0.100 M NH3 is titrated with 20.00 mL of 0.0700 M HCl. Calculate the pH. Kb = 1.76 x 10−5
Consider the titration of 20.00 mL of 0.100 M HBr with 0.150 M KOH at 25...
Consider the titration of 20.00 mL of 0.100 M HBr with 0.150 M KOH at 25 °C. What would be the pH of the solution when 20.00 mL of KOH have been added?
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M NaOH solution....
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M NaOH solution. Calculate the pH at each of the following points. Volume of NaOH added: 0, 5, 10, 25, 40 ,45 , 50 , 55 , 60 , 70 , 80 , 90 , 100
50.0 mL of 0.100 M H2SO3 is titrated with 0.150 M KOH soluion. a) What is...
50.0 mL of 0.100 M H2SO3 is titrated with 0.150 M KOH soluion. a) What is the initial pH of the sulfurous acid? b) How much titrant is needed to reach the equivalence points? What is the pH at each equivalence point? c) What is the pH of the solution after the following points of the titration (with 0.150 M KOH)? (20.0 mL, 33.3 mL, 60.0 mL, 70.0 mL, 100.0 mL, 120.0 mL, 130.0 mL)?
25.0 mL of a 0.100 M HCN solution is titrated with 0.100 M KOH. Ka of...
25.0 mL of a 0.100 M HCN solution is titrated with 0.100 M KOH. Ka of HCN=6.2 x 10^-10. a)What is the pH when 25mL of KOH is added (this is the equivalent point). b) what is the pH when 35 mL of KOH is added.
A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with 0.100 M KOH. The...
A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with 0.100 M KOH. The Ka for the weak acid formic is 1.40 x 10-5. a. Determine the pH for the formic prior to its titration with KOH. b. Determine the pH of this solution at the ½ neutralization point of the titration. c. Identify the conjugate acid-base pair species at the ½ neutralization point.
A 25.0 mL sample of 0.150 M potassium hydroxide is titrated with 0.125 M hydrobromic acid...
A 25.0 mL sample of 0.150 M potassium hydroxide is titrated with 0.125 M hydrobromic acid solution. Calculate the pH after the following volumes of acid have been added: a) 20.0 mL b) 25.0 mL c) 30.0 mL d) 35.0 mL e) 40.0 mL
Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine...
Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine the pH of the solution after the addition of the following volumes of titrant. a. 5.00 mL b. 10.00 mL c. 15.00 mL d. 20.00 mL e. 25.00 mL
Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine...
Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine the pH of the solution after the addition of the following volumes of titrant. a. 5.00 mL b. 10.00 mL c. 15.00 mL d. 20.00 mL e. 25.00 mL
A25.0 mL sample of 0.100 M NaOH is titrated with a 0.100 M HNO3 solution. Calculate...
A25.0 mL sample of 0.100 M NaOH is titrated with a 0.100 M HNO3 solution. Calculate the pH after the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0, 25.1, 26.0, 28.0, and 30.0 of the HNO3.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT